Introduction to Analysis (Pure and Applied Undergraduate Texts) [5th Revised] 0821847872, 9780821847879

Table of contents :
Cover
Title page
Copyright
Preface
Contents
Chapter 0. Preliminaries
Chapter 1. Sequences
Chapter 2. Limits of functions
Chapter 3. Continuity
Chapter 4. Differentiation
Chapter 5. The Riemann integral
Chapter 6. Infinite series
Chapter 7. Sequences and series of functions
Index
Back Cover

Citation preview

Purchased Purchasedfrom fromAmerican AmericanMathematical MathematicalSociety Societyfor forthe theexclusive exclusiveuse useof ofSage SageBinder Binder(bnsgxc) (bnsgxc) Copyright Copyright1998 1998American AmericanMathematical MathematicalSociety. Society.Duplication Duplicationprohibited. prohibited.Please Pleasereport reportunauthorized unauthorizeduse usetotocust-serv@ams cust-serv@am Thank ThankYou! You!Your Yourpurchase purchasesupports supportsthe theAMS' AMS'mission, mission,programs, programs,and andservices servicesfor forthe themathematical mathematicalcommunity. community.

Introduction to Analysis Fifth Edition

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UNDERGRADUATE yVTEXTS • 1

Introduction to Analysis Fifth Edition

Edward D„ Gaughan

American Mathematical Society Providence, Rhode Island

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2000 Mathematics

Subject

Classification.

Primary 26-01.

For a d d i t i o n a l information a n d u p d a t e s on t h i s book, visit www.ams.org/bookpages/amstext-l

L i b r a r y of C o n g r e s s C a t a l o g i n g - i n - P u b l i c a t i o n D a t a Gaughan, Edward. Introduction to analysis—5th ed. / Edward D. Gaughan. p. cm. — (Pure and applied undergraduate texts ; v. 1) Includes index. ISBN 978-0-8218-4787-9 (alk. paper) 1. Mathematical analysis. I. Title. QA300.G34 515—-dc22

2009 2008047387

C o p y i n g and reprinting. Individual readers of this publication, and nonprofit libraries acting for them, are permitted to make fair use of the material, such as to copy a chapter for use in teaching or research. Permission is granted to quote brief passages from this publication in reviews, provided the customary acknowledgment of the source is given. Republication, systematic copying, or multiple reproduction of any material in this publication is permitted only under license from the American Mathematical Society. Requests for such permission should be addressed to the Acquisitions Department, American Mathematical Society, 201 Charles Street, Providence, Rhode Island 02904-2294, USA. Requests can also be made by e-mail to reprint-permission. vi. A U (fl fl C) = (A U £) fl (A U C). Parts (i) through (iv) have very obvious proofs, which will not be supplied here. The proof of (v) follows, and that of (vi) is left as Exercise 3. Proof In proving (v) we shall use Theorem 0.1 to show that A

n (£ u o c (A n £) u (A n o

and (A n £) u (A n o c A n (£ u o , and hence the two sets are equal. Suppose x E A H (£ U C). Then * E A and * E £ U C. If x E £, then since x E A, x E A D £. If * £ £, then since JtEflUC, * E C and also JC E A; hence, x € /4 fl C, Thus, in particular, x E A D B or x EA (1 C. In other words, * E (A H £) U (A fl C). We have shown that A fl (£ U C) C (A fl £) U (A H C). Suppose JC

E (A H £) U (A H C).

Then j c E A H f l or J C E A H C . If x E A H f l , then x E A and x E £; hence, x E A and ^ E £ U C since £ C £ U C. Thus, JC

E A H (£ U C).

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0.1 Sets 5 If x & A D B, then xSADC; hence, x G A and x G C, but C C £ U C, so that * G A H (£ U C). We have shown (A f)B)U

(A (1 C) CA D (B U C),

and the proof is complete. Some of the reasonably easy relations concerning sets given in the exercises will be used later, and you are urged to prove them for your own practice—for the simple reason that a proof makes a theorem much easier to remember. DEFINITION Let A and B be sets. Then the complement of A relative to B, written B \A, is defined to be the set of all objects belonging to B but not to A. In other words, B\A = {*:*G B and* £ A}. Let A = (1, 2, 3, 4, 5, 6} and B = {2, 4, 6, 8, 10}. Then B\A = {8, 10} and A\B = {1, 3, 5}. The following form of what are called DeMorgan's Laws may look a bit cumbersome because of our decision to speak of the complement of A relative to B rather than the complement of a set relative to some implied universal set. However, this path leaves little room for misunderstanding. 03 THEOREM

Let A, B, and C be sets. Then

i. A\(B fl C) = (A\B) U (A\C). ii. A\(B U C ) = (A\B) f) (A\C). Proof Again, the method of proof will be to use Theorem 0.1, showing that the set on the left side of the proposed equality is a subset of the set on the right and vice versa. Although this method of proof will be used quite often in such theorems, it will not always be advertised in the future. (i) Let * G A\(B H C). Then x G A and x & B D C. Since* g B D C, either* $ B or* g C. If * g B, then, since * G A, x G A\B. If * £ C, then since * G A, * G A\C. Hence, * G (A\B) U (A\C). Thus, A\(B

D C) C (A\B) U (A\C).

Suppose now that * G (A\B) U (A\C).

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6 Chapter 0 Preliminaries Then either x E A\B or x E A\C. If x E A \ £ , then x E A and A: £ £, so that x&B (1C; thus, JC E A \ ( £ n C). If x E A \ C , then x E A and jt £ C, so that x £ B fl C; thus JCE

A \ ( £ PI C).

We have shown that (A\B) U (A\C) C A\(B n C). This result, coupled with that of the preceding paragraph, completes the proof of (i). The proof of (ii) is left as Exercise 4. To gain a deeper understanding of Theorem 0.3, let us consider a special case and invent some new notation. Suppose B and C are both subsets of A. If S is a subset of A, let S* denote the complement of S relative to A so that S* = A \ S . In this setting, the theorem states that (B fl C)* = 5 * U C *

and

(B U C)* = B* D C*.

In cruder terms, the complement of the intersection is the union of the complements, and the complement of the union is the intersection of the complements. The notions of union and intersection can be generalized to unions and intersections of more than two sets. But first some notation is needed. DEFINITION Let A be a set, and suppose for each A E A, a subset AA of a given set S is specified. The collection of sets AA is called an indexed family of subsets of S with A as the index set. We denote this by {AA}AeA. DEFINITION

If {AA}AeA is an indexed family of sets, define

fl AA = {x: x E AA for all A E A} AGA

and (J AA = {x: x E AA for some A E A}.

AeA

The reader is invited to generalize some of the previous theorems in light of these new ideas. There is one logical difficulty here that needs to be pointed out. If A is the empty set, then it is easy to see that UAeAAA is empty; however, it is not clear what to expect of Dk&AA^ This could be overcome by insisting that all index sets be nonempty, and this shall be done in some cases. In a context where all sets considered are understood to be subsets of a given set S, the common usage is to let Ae0

where 0 is the empty set. The reason for this discussion is to alert the reader of possible problems that might arise from this situation, not to give a solution. In the event that the index set is the set / of positive integers or a finite subset thereof, some special notation may be adopted. For example, Uwe/Art may also be

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0.1 Sets 7 written U*=|A„, and if S = { 1 , 2 , . . . , &}, we may write Ukn=iAn instead of Unf£SAn. Of course, similar notation may be used for the intersection of a family of sets. •

Example 0.2 Let An = (0, £) for each n E /. Then oo

fl An = 0

n=\

(the empty set).

Let us see why. If x E n~=1An then x E (0, £) for each n E / ; that is, 0 < JC < £ for all n E J. However, if we pick n0 to be any positive integer greater than j , then 0 < ~ < JC and JC £ (0, ~), contrary to our assumption. Thus n~=1Aw is the empty set since no real number can satisfy the conditions for membership. We are assuming some prior knowledge of real numbers, integers, and inequalities, but we will discuss these details later in this chapter. • •

Example 03

Let Bn = (^, 1) for each n E /. Then

oo

U B„ = (0, 1).

n=\

Again, let us see why. Suppose 0 < JC < 1. There is a positive integer m such that ~ < m, hence ^ < JC < 1; that is x E (~, 1). But then, since JC belongs to one set in the indexed family, it must belong to the union of the sets in the indexed family. So (0, 1) C U~=1J3n. On the other hand, if JC E U*=1^„, then JC E Bn for some n E / ; hence, 0 < ln < x < 1. In any case 0 < JC < 1. Thus U^L,^ C (0, 1). Therefore U~=,£w = (0,1). • A generalization of De Morgan's Laws can be stated and proved for indexed families of sets, and such a generalization will be useful later. 0.4 THEOREM (De Morgan's Laws) Let S be a set and {AA}AeA be an indexed family of subsets of 5. Then i. 5\(U AeA A A )=fl, GA (5\A A ). ii. 5\(HAeAAA) = UAGA(5\AA). Proof We will prove (i) and leave (ii) as Exercise 11. Assume JC E S \ (UAeAAA). Then x E S and x € UA6AAA. Therefore x E S and x & AA for all A E A. Therefore, JC E S\AA for all A E A. This means that JC E HAeA(5'\AA). On the other hand, if ^ £ nAGA(S\AA), then JC E S\AX for each A E A. Thus JC E 5 and JC £ AA for each A E A. This implies that JC E S and x & UAeAAA. So JC E S\(UA6AAA). We have shown that S\( |J A,)c \AeA

/

fl (S\AX)

AeA

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8 Chapter 0 Preliminaries

and

fl (S\A A )C5\(U AX Therefore, the two sets are equal. •

Example 0.4 Using Example 0.2 and De Morgan's Laws, we see that

U.K*;))-*^.!*;;)-™-*-

0.2 RELATIONS AND FUNCTIONS The notion of a function is basic to the study of mathematics and you have used it many times in your previous mathematical experiences. The usual informal definition states that a function is a correspondence that associates with each element in the domain of the function a unique element in the range of the function. In this section we offer a formal and more precise definition of function and explore some of the properties of functions. The idea of a function is very simple. It is a means whereby an input determines an output in some prescribed manner. Most of your experiences involved functions defined by a formula, a rule, a graph, a table, or a machine such as a calculator or computer. But the basic requirement is that given the input, there is no ambiguity about the output. If you feel very comfortable with the concept of function, you may want to read this section somewhat casually. Otherwise, you may want to follow the discussion very carefully. To give a precise definition of function we need to define ordered pair, Cartesian Product, and relation. DEFINITION The ordered pair (x, y) is defined to be the set {x, {x, y}}. Our purpose is to define the ordered pair (x, y) in such a way that the order in which the objects x and y appear is important. As a result we refer to x as thefirstcoordinate and y as the second coordinate of the ordered pair (x, y). The next theorem justifies this language. We leave the proof as an exercise. It is straightforward, but a little tedious. 0.5 THEOREM (x, y) = (w, v) if and only if x = u and y = v. Theorem 0.5 simply states that two ordered pairs are equal if and only if the first coordinates are the same and the second coordinates are the same. This is neither a new nor startling result, but is included for the sake of completeness. DEFINITION If A and B are sets, define the Cartesian Product of A and Bt written A X Bf to be the set of all ordered pairs (a, b) where a E A and b E B. Thus AX B = {(a, b): a E A and b E B).

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0.2 Relations and Functions 9

• Example OJ Let A = {1,2,3} a n d £ = {2,4, 6}. Then A XB= (1, 6), (2, 2), (2, 4), (2, 6), (3, 2), (3, 4), (3, 6)}.

{(1, 2), (1,4), •

A note of caution is in order here. Recall from Section 0.1 that we used the notation (a, b) to denote the interval {x: a < x < b). Now we are using (a, b) to denote the ordered pair with first coordinate a and second coordinate b. The good news is that the danger of confusion is more imagined than real. When we write (a, b), it will be clear from the context whether we are referring to an open interval or an ordered pair. DEFINITION A relation is a set of ordered pairs. A function F is a relation such that if (JC, y) E F and (x9 z) E F, then y = z. If F is a function, then the domain of F, written dom F, is defined to be dom F = {x : there is y such that (x, j ) E F} and the image of F, written im F, is defined to be im F = {y : there is x such that (x, y) E F}. Examine the definition of function and reflect on the meaning. The function F is a set of ordered pairs such that for each x E dom F, there is exactly one pair (JC, y) E F. It is customary to denote y by F(x). It is useful to think of F in a dynamic way. Input x from dom F and F produces an output F(x). Suppose that F is a function with dom F = A and im F C B. We say that F is a function from A to B and B is called the codomain of F. We write this symbolically as:

F:A~»B. One reason for specifying the codomain of a function rather than the image is that determining the image is sometimes very difficult. • Example 0.6 Functions may be defined in a variety of ways. A common means is to specify the domain and give a rule or formula. For example, define / : / - > / as follows: f(n) = 2 / i - l for each n E / . Notice that the rule is unambiguous, that is, given n EJ, there is no doubt about f(n). Also notice that the domain is specified. • Now comes confession time. Remember back in Section 0.1 when we defined an indexed family of sets? We really were defining a function. An explanation is in order. Let A and S be sets, with P(S) the set of all subsets of S. Let A be a function with domain A and codomain P(S). Then the indexed family {AA}AeA is simply the function A : A —> P(S) where we write AA instead of A(A). In the discussion of intersection of indexed families of sets, the question arose concerning the result when A is the empty set. This raises the question, "Is the empty set a function?" The answer is yes, but it

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10 Chapter 0 Preliminaries

is not a very interesting function. If this bothers you, simply include in the definition of function the requirement that a function must be nonempty. Consider now the functions f(x) = x3, x E R, and g(x) = *¥x, x E R. There is a special relationship between / and g that we will discuss shortly. For the moment, notice that /(2) = 8 and g(8) = 2. This means that g "undoes" what / "does" and vice versa. You have seen other pairs of functions with this property before, and we will examine the idea in detail. DEFINITION Let S be a relation. Then the converse of S, written S, is defined by S=

{(x,y):(y,x)SS\.

It would be interesting to discover under what conditions the converse of a function is also a function. Suppose F is a function with (JC, y) and (x, z) both belonging to F. In order that F be a function, it must be true that y = z. Translating back to Ft for all (y, JC) and (z, x) in F, it must be true that y = z; or, in other words, if F(y) = F(z), then j = z. This idea is important enough to deserve a special definition. DEFINITION A function / is 1-1 (pronounced "one to one") if and only if, for all y, z in the domain of /, f(y) = /(z) implies y = z. In essence, this says that a 1-1 function is one that assumes each value in its image exactly once. The next theorem is another "if and only if" theorem that takes the form: "Property A is satisfied if and only if Property B is satisfied.'' It has become accepted practice in mathematics literature to use "iff" as an abbreviation for "if and only if." Thus, the theorem could be written: "Property A is satisfied iff Property B is satisfied." 0.6 THEOREM Let F be a function. Then F (the converse of F) is a function iff F is a 1-1 function. If F is a function, then dom F = im F and im F = dom F. Proof Assume F is a function. Suppose x and y belong to the domain of F with F(x\ = F(y) = z- Then (JC, z) E F and (y,z) E F so that (z, y) E F and (z, *) E F. Since F is a function, y = JC. Thus, if F is a function, then F is 1-1. Now assume F is 1-1. Let (w, w) E F and («, u) E F. To show that F is a function, we must show that w = y. By the definition of F, (w, u) E F and (y, w) E F, making M = F(w) = F(y). By assumption, F is 1-1; hence w = v. The facts that im F = dom F and dom F = im F are immediate. If F is a 1-1 function, we shall write F" 1 in place of F and call F" 1 the inverse of F. Furthermore, if F is 1-1, then F = F is a function, so F = F~l is also 1-L Suppose f ;A-+ B and T C A, Define f(T)= {/«:* e r » . /(J) is called the /nw#? of T under /. In particular f(A) = im /. If C C 5, define r l ( Q = {aEA:/(a)EC}.

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0.2 Relations and Functions 11 f~\C) is called the inverse image of C under /. Note that this definition is given without any assumption that / is 1-1, so it is not assumed here that / is a function. I f / : A — » f H s a l - l function, / is sometimes called an injection or an injective function. When im / = B, we say that / is a function from A onto B and call / a surjection or a surjective function. If / : A —> B is both 1-1 and onto, / is called a bisection or a bijective function. DEFINITION If / : A -* B and g : B -> C, the composition of g by /, written g ° /, is defined to be the set {(x, y): there is w E B such that (x, w) E / and (w, j) E g}. 0.7 THEOREM If / : A -> £ and g : fi -» C, then g / is a function and go/:A-^C. Proof Suppose (x, M) E g ° / and (x, v) £ g° f. We must show that u = y. By the definition of g ° / there are vt^, w2 E £ such that (x, Wj) E /, (wb w) E g, (x, w2) E /, and (w2, 0) E g. Since / is a function, w} = w2; since g is a function, w = u. Thus g ° / is a function. Let / : A —» £ and g : f* -» C. If x E A, then there is a ;y E B such that (x, y) E /, or y = /(x); also, there is z E C such that (j, z) E g, or z = g(y). This means that (x, z) E g © / and z = g ° /(x). But z = s(y) = *(/W), so

(*°/)W = *(/W) and we see that the composition of g by / is accomplished by ''following'' / by g. Also, it is clear the dom (g ° /) = dom /. Though it is discussed at length here, the composition of two functions is not a new idea. For example, the function h(x) = sin x2 is the composition of the function g(x) = sin x by f(x) = x2, so that h = g° f. Note that / ° g is the function whose value at x is (sin x)2; hence f°gizg0f. If S is a set, denote by l s the function from S onto S defined by ls(s) = s for each s E S. Now, if / is a 1-1 function from A onto B, then we have seen that f~l is a 1-1 function from £ onto A. Upon examination of the definition of / ~ \ it becomes clear t h a t / o / - 1 = l B a n d / - ! o / = iA. •

Example 0.7 Let S = {x:x E /? and x # - 1 } . Define / : S -> /? by 2x - 1 2a - 1 26-1 /(*) = — — r . First of all, if f(a) = f(b)9 then — — - = /(a) = /ffe) = — — - . A x+1 a+ 1 0+1 bit of algebra shows that a = b, hence / i s 1-1. Let us find im / and, in the process, we will discover / " l . The number y E im / if and only if there is x E dom / = S such

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12 Chapter 0 Preliminaries

that fix) = v—that is, y =

7-. Solving for x, we obtain x = . This means x +1 2- y that if y # 2, there is JC E S such that /(JC) = y. Thus, im / = R \ {2}. Also, this means y+ 1 J

W

/

J

thatr'W = | ^ .

• Example 0.8 Sometimes we know a function has an inverse, butfindit difficult to obtain a formula for that inverse. Consider the function / : [0, 2] ~» R defined by f(x) = x3 - 3x2 for all x E [0, 2]. A bit of calculus (the Mean Value Theorem) shows us that / is 1-1. However, we would be hard pressed tofinda formula for f~l.

0.3 MATHEMATICAL INDUCTION AND RECURSION In what is to follow, we will have need of a technique of proof called mathematical induction. In some cases, the use will be rather informal, as in the case of showing that every polynomial function has a limit at every point., On the other hand, in less obvious applications, such as Taylor's Theorem in Chapter 5, the approach will be formal and very precise. Recursion will be used in the definition of certain functions and sequences defined on /. Usually a few initial values of the function are given, and then the value at n is given in terms of the values of the function at k for some k < n. The basis for mathematical induction is the well-ordering principle, which we assume without proof. WELL-ORDERING PRINCIPLE Every nonempty subset of / has a smallest member. 0.8 THEOREM (Principle of Mathematical Induction) If P(n) is a statement containing the variable n such that i. P(l) is a true statement, and ii. for each k E J, if P(k) is true, then P(k 4- 1) is true, then P(n) is true for all n EJ. Reflect on the implication of Theorem 0.8. Basically it says that if P(l) is true, and if one can establish that P(k + 1) is true from the truth of P(k), then P(n) is true for every positive integer n. Intuitively, this means that from the truth of P(l) we can infer that P(2) is true, and from the truth of P(2), we can infer that P(3) is true, and so on. The proof of this theorem is ourfirstexample of a proof by contradiction. Usually a theorem is stated in the form "If A is true, then B is true." To prove such a theorem by contradiction, one assumes that A is true and B is false and then shows that this leads to a logical contradiction, often by showing that A is false. Consequently, the assumption "A is true and B is false" can't be true, so the theorem must be true.

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03 Mathematical Induction and Recursion 13 Proof Assume that P(n) is a statement about the variable n such that i. P{\) is true, and ii. for each k E /, if P(k) is true, then P(k + 1) is true. In addition, assume that P(n) is not true for some n E /. (This is the "A is true, B is false" part of the proof by contradiction.) Since P(n) is false for some n E /, the set S = \n\P{n) is false} is a nonvoid subset of/. As such, by the well-ordering principle, S has a smallest member; call it n0. Since P(l) is true, nQ > 1. Since «0 is the smallest member of 5, P(n0) is false and P(n0 - 1) is true. But remember, if P(fc) is true, then P(k + 1) is true! So, if we let k = n0 - 1, then />(£) is true, but k + 1 = «0 and P(& + 1) is false. This is a contradiction. Hence, by the method of proof by contradiction, the theorem is true. In the pages to come, you will see more proofs by contradiction, so now is a good time to examine the process by reading the proof several times for maximum understanding. •

Example 0.9 The following identity will be needed in Chapter 6: For n E /,

1 -

1 + x + x2 + • • * + xn =

JCW+1

1- x

if x * 1.

We will use mathematical induction to verify this identity. Let P(n) be the statement 1 - x n+1 1 + x + x2 + • • • + xn = — if x * 1. 1- x a. First check to see that P(\) is true. If n = 1, the left member of this identity is 1 - x2 1 + x and the right member is = 1 + xiix # 1. Thus, P(l) is true. 1—x b. Now assume that P(k) is true and try to show that this implies that P(k + 1) is true. So we have 1 - xk+l 2 k 1 + x + x + •. • 4- x = — if x * 1. so 1 + x + x2 + • • • + xk + x*+1 =

— + **+1 1—x (because we assumed P(k) is true) k+l + xk+l(l - x) = 1 ~ x 1 - x k+2 1- x = * i f * * 1. 1- x

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14 Chapter 0 Preliminaries

This last equality asserts that P(k 4- 1) is true. Thus, from the truth of P(&), we can prove that P(k + 1) is true. By the principle of mathematical induction, P(n) is true for all n E /. This was our goal. • Under some circumstances, we need what is sometimes referred to as the second principle of mathematical induction. 0.9 THEOREM (Second Principle of Mathematical Induction) Suppose that P(ri) is a statement concerning the variable n. If i. P(l), P(2),..., P(m) are true, and ii. for k E /, k ^ m, if P(i) is true for all i E / such that 1 < / < £, then P(k + 1) is true, then P(n) is true for all « G / , Proof We leave the proof as Exercise 23. The proof is very similar to that of Theorem 0.8 and uses the well-ordering principle. The function defined in Example 0.10 is defined by recursion; that is, some initial values of / are given—in this case, /(l) and f(2)—and then, for larger n, f(n) is given in terms of the values f(k) for some k < n. In this case f(n) = -

. We will use the second principle of mathematical induction to

prove that f(n) = 1 + (-^)n~l for all « £ / , •

Example 0.10 Define / : / -> R as follows: /(l) = 3, /(2) = - ,

and

for n > 3 /(n) =

.

We want to prove that f(n) = 2 + (-5)'1"1 for all n E /. To do so, we will use Theorem 0.9. Let P(n) be the statement that f(n) = 2 + (-5)j i - i a. We will show that P(l) and P(2) are true. For n = 1 the formula gives 2 + (™j

= 2 + 1 = 3 = /(l).

For n = 2, the formula gives

(4)

2 + i ~ | = 2 - i = | = /(2). So, forn = 1 andn = 2,/(n) = 2 + (-i)" - 1 .

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0.3 Mathematical Induction and Recursion 15 b. Assume now that /(i) = 2 + (-5) 1 " 1 for 1 < / ^ k. Then,

w+1) .ffi±f -i»4[ 2+ (4)- +2+ (_l)"j 4+rl~l

1=2+1-^

Thus, the formula holds for w = k + 1. By the second principle of mathematical induction, f(n) = 2 + ( - | ) n _ 1 for all

nEJ.

m

The proofs of both versions of mathematical induction depend on the well-ordering principle of the set of positive integers. Both the well-ordering principle and the two principles of mathematical induction can be modified easily as follows. WELL-ORDERING PRINCIPLE (Modified) If S is any nonempty subset of Z (remember, Z is the set of all integers) such that S has a smallest member, then any nonempty subset of S has a smallest member. Proof Suppose S is a nonempty subset of Z and s0 is the smallest member of S. Let A be a nonvoid subset of S. If A C / , then the first version of the well-ordering principle gives the result. If A is not a subset of / , then s 0 < 1. Let T = {% so + 1, s0 + 2 , . . . , 0}. Then A D T is a nonvoid finite set, and the smallest member of A is the smallest member of A C\ T. If you reexamine the proof of Theorem 0.8, you will see that the argument relied in part on the fact that, if P(n) was false for some n, then there was a smallest such n and, since P(l) was true, that particular n was greater than 1. Using the modified well-ordering principle, we can prove the following modification of Theorem 0.8. 0.10 THEOREM (Modified Principle of Mathematical Induction) P(n) is a statement concerning the variable n. If

Suppose that

i. for some n0 E Z, P(n0) is true, and ii. for each k €E Z, k ^ n0, if P(k) is true, then P(k + 1) is true, then P(n) is true from all n G Z, n ^ n0. Proof

The proof is left as Exercise 27.

The following example shows how Theorem 0.10 might be applied. •

Example 0.11 Consider the statement P(n): 2n + 1 < 2n. Try a few values of n: n = 1 2 - l + l = 3 i £ 2 = 21 F(l) is false. n = 2 2 - 2 + 1 = 5 ^ 4 = 2 2 P(2) is false. « = 3 3 - 2 + l = 7 < 8 = 2 3 Aha! F(3) is true.

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16 Chapter 0 Preliminaries

So we will apply the modified version of mathematical induction with n0 = 3. a. F(3) is true (see above). b. Assume k > 3 and P(k) is true. Then 2(k + 1) + 1 = 2k + 1 + 2 < 2* + 2 < 2* + 2* = 2*+1. Thus,P(£ + 1) is true. Therefore, POO is true for all « > 3.

•

Exercise 28 asks you to prove a modified version of Theorem 0.9.

0.4 EQUIVALENT AND COUNTABLE SETS Consider now the sets A = {1, 2, 3, 4, 5} and £ = {2, 4, 6, 8, 10}. It is clear that A ± Bf and in fact A (t B and £ 0, a and b are relatively

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20 Chapter 0 Preliminaries

prime (that is, a and b have no common divisors greater than 1), and, if a = 0, b = 1. We define a function / : S —> Z X / as follows: f(r) = / I - J = (a, &) where 7 is the unique representation \bj b of the rational number r. This function is 1-1 from S onto a subset of Z X /, and, by Theorem 0.16, Z X / is countable; hence, any subset of Z X / is countable by Corollary 0.15. Therefore, since S is equivalent to a subset of Z X /, then S is countable. • There are many ways of combining countable sets to obtain countable sets. Theorem 0.16 gives one such way, and the next theorem gives another method. This theorem will be useful in Chapter 2. 0J7 THEOREM Let S be a nonempty subset of /. Let {As}seS be an indexed family of countable sets. Then USBSAS is a countable set. Proof Since each As is countable, for each s E S, there is a 1-1 function fs: As -* / . If x E UssSAs, then there is a smallest member m of S such that x E Am. We will attempt to define a function / : UJ65A^ -> / X / as follows: f(x) = (m, fm(x)% where m is the least integer in S such that x E Am. Since x E U^eSAs, it must belong to Am for some m& S; hence there is a smallest such m by the well-ordering principle. Thus / is a well-defined function—that is, the rule forfindingf(x) is unambiguous. We now need to show that / is 1-1. Suppose fix) = f(y) = (m, 11). Thus, xGAm and y E Am\ hence, n = fm{x) = / m (j). Because fm is 1-1, x = y. Thus, / is a 1-1 function. We have constructed a 1-1 function / from Uje5As onto a subset of / X / ; hence, UJ€SA^ is a countable set. Our preoccupation with countable sets would seem bizarre if all infinite sets were countable. However, there are many uncountable sets. The set of real numbers is a familiar example, although you may not be familiar with the fact that R is uncountable. In Exercise 37 you are asked to prove that, for any set A, P(A)—the set of all subsets of A—is not equivalent to A. Hence, the set of all subsets of J is another example of an uncountable set. There are several ways to prove that R is uncountable. In one of the projects at the end of Chapter 1, you will have an opportunity to provide such a proof. In anticipation of that project, we next give an example of a set equivalent to R. • Example 0.13 We will show that R is equivalent to (0, 1). The easiest way to accomplish this is to use a function familiar to anyone who has studied trigonometry. Define / : (0, 1) ~» R by f(x) = tan (irx - f). This gives a 1-1 function from (0, 1) to/?. • See Project 0.1 at the end of this chapter for other sets equivalent to (0,1).

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0.5 Real Numbers 21

0.5 REAL NUMBERS You probably have a good knowledge of the set of rational numbers, the order relation of that set, and the operations of addition and multiplication. However, the set of rational numbers is not sufficient to support all of mathematics. The following theorem shows that an equation as simple as x2 = 2 has no solution in the set of rational numbers. 0.18 THEOREM

There is no rational number whose square is 2.

Proof Suppose there are positive integers p and q such that 2 = (p/q)2. Assume further that p and q are relatively prime—that is, their greatest common divisor is I. Thus, we have 2q2 = p2, so p is even (since p odd implies p2 odd). We may write p = 2r where r is an integer. Then 2q2 = p2 = 4r 2 , so q2 = 2r2, and hence q is even. Thus, both p and q are even, contrary to the assumption that p and q are relatively prime. Rather than construct the set of real numbers, we shall assume the existence of that set and postulate the properties that we will need. For those who wish a careful and rigorous development of the real numbers, see Edmund Landau's Foundations of Analysis, 2d ed. (New York: Chelsea, 1960). We assume the existence of a set Rt called the set of real numbers, that satisfies the following axioms. There are functions + :R X R ~-> R and *:R X R -» R and a relation < on R such that, for all x, y, z E Rf we have 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. II.

(x 4- y) 4- z = x + (y 4- z); (x • y) • z = x • (y • z). x + y = y 4- JC; x • y = y • JC. *-(y + z) = (*-;y) + (*-z). There is a unique element 0 E R such that 0 4- x = JC for all x E /?. For each x E R, there is a unique y E R such that JC 4- y = 0, and we write y = -x There is a unique element 1 E Z? such that x • 1 = * for all JC E R and 0 # 1. For each JC E /? with JC =£ 0, there is a unique element y E J? such that JC • y = 1, and we write y = JC"1 or y = £. JC < y implies JC + z < y + z. JC < y and y < z implies JC < z. For JC, y E R, exactly one of the following is true: x < y, y < x, or JC = y. JC < y and z > 0 implies JCZ < yz.

This list of properties assures us that the system (/?, 4-, •, M). A set is bounded if it is bounded both from above and from below. If S C Rf a real number M is an wpper fojmd (lower bound) for 5 if for all JC E 5, JC ^ M (x ^ M). If a nonvoid set S is finite, then S is obviously bounded; and, in fact, among the members of S there must be a largest, which we denote

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22 Chapter 0 Preliminaries

by max S, and a smallest, which we denote by min S. For an infinite set S, it is not necessarily true that 5 has either a largest or a smallest number. Let S be a set of real numbers bounded from above. A real number a is a least upper bound for S if a is an upper bound for S having the property that, if b is also an upper bound for S, then a ^ b. If S is bounded from below, then a real number a is a greatest lower bound for S if a is a lower bound for S having the property that, if b is any lower bound for Sf then b < a. It is immediately clear that if 5 has a least upper bound a, then it is unique; we write a = l.u.b. S or a = swp S. The abbreviation swp is from the word supremum, often used as a synonym for least upper bound. Likewise, if S has a greatest lower bound c, then it is unique; we write c = g.l.b. S or c = //i/S. //i/ comes from inflmum, a synonym for greatest lower bound. The next property of the set of real numbers is very important to the analyst. It is called the least upper bound property. 12. Every nonempty set of real numbers that is bounded from above has a least upper bound. Axioms 1-12 tell us that (/?, + , • , < ) is a complete ordered field. Axioms 1-7 are the axioms for a field, axioms 8-11 indicate that the order relation interacts with addition and multiplication in the proper manner, and axiom 12 is required that the ordered field be complete. The set of rational numbers with +, •, and < is an ordered field, but not a complete ordered field. The set {x : x is a rational number and x2 < 2} has no least upper bound in the set of rational numbers. The proof of this statement is left as Exercise 43. However, before you tackle that exercise, you will need some more operating machinery. Axioms 1-12 were designed to be a minimum set of axioms from which we can prove other relations that will be useful later. The next theorem offers some of those relations. 0.19 THEOREM i. ii. iii. iv. v.

Let x, y, and z be any real numbers. Then

If x < y, then -y < —x. 0 < 1. If 0 < x < y, then 0 < j < ~. If x < y and z < 0, then yz < xz. x2 > 0.

Proof (i) Assume x < y. Then x+(—x—y) < y+(—x—y) by axiom 8. Removing parentheses and simplifying yields — y < —x. (ii) This statement may seem out of place and perhaps ridiculous, but the result is needed in the proof of (iii). Besides, this statement should be a result of the axioms. By axiom 10, exactly one of the following holds: 0 = l , 0 < l , o r l < 0 . We know that 0 ^ 1, so consider the possibility 1 < 0. Then by (i) above, 0 < - 1 . But then, by axiom 11, 0 • (-1) < ( - 1 ) ( - 1 ) , or 0 < 1. This contradicts the assumption 1 < 0. The only remaining case is 0 < 1. Since one of the three statements—0 = 1, 0 < 1, or 1 < 0—must be true, it is the case that 0 < 1.

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0.5 Real Numbers 23

(iii) Suppose 0 < x < y. Now, by axiom 10, £ = 0, 7 < 0, or j > 0. Since x • £ = 1, we rule out £ = 0. Suppose that 7 < 0. Then, since x > 0, 1 == x • j < jt • 0 = 0, which contradicts (ii). Therefore, \ and j are both greater than zero. But then

• 0 \ - j c i f J C < 0.

We call |*| the absolute value of x. Note that |JC| = Vx 5 and x ^ |JC| for all x £ R. 025 THEOREM Let a and b be any real numbers. Then i. ii. hi. iv.

|afr| = |*l!*l If €> 0, then \a\ < €iff - € < a < €. |tf + fe|^|tf| + |fc| (the triangle inequality) ||a| — |6|| ^ \a - b\

Proof i. Let a and b be any two real numbers. Then

\ab\ = VW

= V 7 F = V ? V F = \a\\b\.

ii. Suppose -e < a < e, e > 0. If a > 0, then \a\ = a < €. If a ^ 0, then \a\ = -a and -e < a; hence \a\ = -a < e Thus, -€ < a < 6, € > 0

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Exercises 27 implies that \a\ < e. If \a\ < e and a > 0, then - e < 0 < a = \a\ < e. If a < 0, then |a| = - a < e; hence - e < « < 0 < 6. Thus, |a| < € implies that —e 2, /(h) = V3 + /(n - 1). Prove that f(n) < 2.4 for all n G /. You may want to use your calculator for this exercise.

20. 21. 22. 23. 24.

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Project 0.1 29 26. Define f:J-+J by /(l) = 2, /(2) = - 8 , and, for n ss: 3, /(n) = 8/(n - 1) - 15/(« - 2) + 6 • 2n. Prove that, for all n E J, /(*) = - 5 « 3* + S^ 1 + 2',+3. 27. Prove Theorem 0,10. *28. Prove the following modified version of the second principle of mathematical induction: Let P(n) be a statement for each n E Z. If (a) P(n0), P(n0 + 1 ) , . . . , P(m) is true, and (b) for k Sr m, if />(#) is true for % < i < &, then />(& 4- 1) is true, then P(rt) is true for all n > %, n & Z. 29. Define /(w) as follows for « 6 Z, « > 0, /(0) = 7, /(I) = 4, and, for it > 2, /(«) = 6/(» - 2) - f(n - 1). Prove that f(n) = 5 • 2n + 2(-3f for all n E Z, w ^ 0. 0.4 EQUIVALENT AND COUNTABLE SETS 30. Prove Corollary 0.15. 31. Find a 1-1 function / from / onto S where 5 is the set of all odd integers. 32. Let P„ be the set of all polynomials of degree n with integer coefficients. Prove that Pn is countable. (Hint: A proof by induction is one method of approach.) 33. Use Exercise 32 to show that the set of all polynomials with integer coefficients is a countable set. 34. Prove the following generalization of Theorem 0.17: If S is a countable set and |/!,v}%tes is an indexed family of countable sets, then UsesA5 is a countable set. 35. For eachp E P„, define B(p) - \x :p(x) = 0}. Prove that UP€PH B(p) is countable. 36. An algebraic number is any number that is a root of a polynomial equation p(x) = 0 where the coefficients of p are integers. Show that the set of algebraic numbers is a countable set. 37. For a set A, let P(A) be the set of all subsets of A, Prove that A is not equivalent to P(A), (Hint; Suppose / : A -> P(A) and define C = {x: x 6 A and x € fix)), Show C $ im /.) 38. Let a, b, c, and d be any real numbers such that a < b and c < d, Prove that [a, b] is equivalent to [c, d\. (Hint: Show that [a, b] is equivalent to [0, 1] first.) 0.5 REAL NUMBERS *39. If x < y, prove that x < ^ — - < y. *40. If JC ^ 0 and v > 0, prove that Vxy =s ^ — - . (Hint: Use the fact that (Vx - Vyf > 0.) *41. if o < a < b, prove that 0 < a2 < b2 and 0 < Va < Vb. .**,. ,, , x a , x x +a a 42. If xf y, ar and b are greater than zero and - < - , prove that - < < ~-. J y b y y +b b 43. Let A = (r: r is a rational number and r2 < 2}. Prove the A has no largest member. (Hint: 2- r2 2 If r < 2, and r > 0, choose a rational number 8 such that 0 < 8 < 1 and 8 < . 2r + 1 Show that (r + S)2 < 2.) *44. If JC = sup 8, show that, for each e > 0, there ha S S such that x — € < a ^ x. *45. If y = inf S, show that, for each €> 0, there is a E S such that y < a < y + €.

PROJECT 0.1

,

.

The purpose of this project is to show that the open interval (0, 1) is equivalent to the closed interval [0,1], In the process we will discover that both intervals are equivalent to [0, 1) and (0, 1]. It is then easy to generalize to any interval [a, b] with a 0 there is a positive integer N such that for all n > N we have | an — A \ < e. It is worthwhile to point out that the choice of N may depend upon the choice of e. Let us return to the sequence {^},T=i- Intuition leads us to believe that this sequence should converge to zero. If this conclusion is correct, then for € > 0, there is N such that, for n ^ N, \a„ - 0| = \Ln - 0| = i < e. For example, if e = .025, then, for / i > 4 1 , | a r t - 0 | = J ^ j r < .025. Thus, for e = .025, N = 41 satisfies the conditions of the definition. For e = .00025, one may take N = 4001. Note that, if N is such that for n > N, \a„ ~ A\ < e and N < M, then, for n ^ M, \an - A\ < e. Hence, if {an\n=\ converges to A, then for each e > 0, there are many, in fact infinitely many, positive integers N such that for n ^ N, \an - A\ < €. However, to prove that {an}™=l converges to A, it suffices to show that for each e > 0, there is at least one number N with the desired property. Follow along to see that {£}~=i converges to 0. Choose e > 0. Let N be an integer larger than £. So if n S: JV, we have ~ < £ < e. This means that if an = { and A = 0, then for n ^ N, \an ~ A\ = $ < e. Therefore the sequence converges to zero. By the way, do you see now why we chose N = 41 when 6 = .025? In order to digest this new idea completely, we look for other ways of expressing the notion. Recall that, if a and b are real numbers and e > 0, then \a - b\ < e iff h - e < a < b + €. Thus, { 0. Let N be any positive integer greater than ~. Then for n ^ iV, we have

i-o n

1 1 = - < — 0. There is a positive integer N such that, if n, m >. N, then A — e < an < A + e and A-€ 0. There is a positive integer N such that n ^ N implies \an - A < f. (The choice of f is not a mere whim. We observed previously that the difference between a„ and am was less than twice the original choice of e.) Now, if m, n ^ N, then \an ~ A | < § and \am - A\ < f; hence, \N implies that \an - am\ < €. Since aw = sy and am = sk for some j and & among { 1 , . . . , r} and e was chosen to be the least distance between distinct members of the range, we must have an = am for n, m^ N. Thus, the sequence is constant from some point on and hence converges (see Exercise 11). If the range of a Cauchy sequence is infinite, then, by Theorem 1.4, the range is an infinite bounded set and hence, by the Bolzano-Weierstrass Theorem, has an accumulation point. We have been searching for a point where the sequence "piles up,'* and this accumulation point should be it. 1.7 THEOREM Every Cauchy sequence is convergent. Proof Let {a„}~=1 be a Cauchy sequence. By the preceding remarks, if the range is finite, the sequence is constant from some point on; hence it converges. Suppose the range is infinite—call it S. By Theorem 1.4, S is bounded; hence, by the Bolzano-Weierstrass Theorem, 5 has an accumulation point—call it a. We shall prove that {an}„=l converges to a. Choose e > 0. Since (a - f, a + f) is a neighborhood of a, it contains infinitely many members of the set S. Since {aw}~=1 is Cauchy, there is a positive integer N such that nfm*±N implies \ 0. There is a positive integer Nx such that if n ^ Nu then | N, we have n ^ iVl5 so \an - A\ < § and « > iV2; hence, \bn - B| < §. Thus, \(an 4- bn) - (A + B)| = |(an - A) + (6* - B)\

Thus, {a„ + &„}~=i converges to A + B. Proofs like the preceding do not stem from divine inspiration. You begin with what is needed, \(an + bn) - (A + B)\ < e, and then you work backward to find the appropriate choice of N. Although proofs are not written this way, they are conceived in this manner. Let us follow this process for the product \anbn}n^\ of two sequences to see how the process works. Suppose \an}™=i converges to A and {fc„}~=i converges to B. In light of previous discussion, it seems reasonable to try to prove that {anbn}„=x converges to AB. Now for the scratch work: \anbn - AB\ = \anbn - anB + anB - AB\ = \an(bn - B) + B(an - A)| ^\an\\bn -B\ + \B\\an~-A\. Since {an}~=i converges to A and {bn)n=\ converges to Bt we can make \an - A\ and \bn - B\ small. The constant \B\ poses no problem, but \an\ depends on the choice of n. Now it is appropriate to recall an earlier theorem—namely, that every convergent sequence is bounded. Thus, although \a„\ depends on the choice of n, it can't be very large. Suppose \an\ ^ M for all «. Then, if e > 0 is chosen, we wish to force the following: |a,116, - B\ + \B\\an - A\ * M\bn - fl| + \B\\an - A\ < e. This is satisfied if

lb

"-Bln)n=\ fails to converge. Exercises 25, 26, and 27 illustrate this point. The problem with division must be approached with some caution. First of all, consider {an}~=1 converging to A and A}«=i converging to B. We wish to consider the sequence

ULi and might be led to conjecture that this converges to f. For this even to make sense, we must insist that B ^ O and that bn # 0 for all n. Let us proceed with these facts in mind. bn

B

anB - AB + AB - bnA

anB - bnA KB

KB
0. By Lemma 1.10, there is a positive real number M and a positive integer Nt such that \bn\ ^ M for all n^Nx. Then Me

>0.

1 + 1*1

(See the paragraphs preceding Lemma 1.10 to understand the reason for this choice of €'.) There is a positive integer N2 such that, for n ^ N2, \an — A| < e', and a positive integer N3 such that n > JV3 implies |6n - fi| < e'. Let iV = maxliV,, N29 N3}. For n > N, \bn ~ B\ < c'9 \an -A\< e'f and \bn\ > M. Thus, anB-\ KA anB - AB + AB - bnA KB KB a„- A \A\\bn-B\

52 _ A bn B

+

K 0, > rn + 1— n V r t + 1 - Vn = Vn + 1 + Vn

1 Vn + 1 + Vn'

Now use us refer to some of the homework exercises rather than attack this problem directly. By Exercise 28, \ —7= \ 0
an+x for all positive integers n. A sequence is monotone iff it is either increasing or decreasing. 1.16 THEOREM

A monotone sequence is convergent iff it is bounded.

Proof Suppose {an} £= i is a monotone sequence that is bounded; for definiteness, suppose the sequence is increasing. The set {an : n = 1, 2 , . . . } is bounded, so let s = mp{an ; n = 1, 2 , . . . } . It will be shown that {an)~=\ converges to s. Choose e > 0. Since s is the least upper bound of {an : n = 1, 2 , . . . } , s — e is not an upper bound; hence, there is n0 such that s - e < ano. Now, for n > n0i s - e < anQ ^ an < s < s + e\ hence f aw}£=i converges to s. For the case in which {an}n=\ is decreasing, let s — mi\an : n — 1, 2 , . . . } , and the details are similar (see Exercise 37). If the sequence is convergent, then by Theorem 1.2 it is bounded. In keeping with our philosophy of examining examples to illustrate theorems and to show the need for the hypotheses, we shall consider several sequences. First, the sequence {n}*=1 is indeed monotone, but it does not converge because it is unbounded. The sequence {0, 1, 0 , . . . } is bounded, but the theorem does not apply, because it is not monotone; indeed, it fails to converge. •

Example 1.6 5! = V 2 ,

Consider the sequence {sn}~=i defined as follows: sn = \/l

+ Vs^Z

for

n = 2, 3 , . . . .

We shall prove that this sequence is increasing and bounded by 2 and hence, by the preceding theorem, is convergent. Since this sequence is defined recursively, we will use induction to prove the conjecture stated above. Clearly, st ^ 2 and s2 = V 2 + i/2 > V 2 = sx. We wish to show that sn ^ 2 and sn+x > sn for all n\ the preceding statement verifies the truth of this for n - 1. Assume the condition is satisfied for n = r. Then sr+1 = V 2 + V ^ ss V 2 + V 2 < V 2 + 2 = 2 and

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52 Chapter 1 Sequences hence, by the principle of induction, the inequalities hold for all n. By use of Theorem 1.16 we are assured that the sequence converges; call the limit L. Now the subsequence {sn~\}n=2 also converges to L, and since sn > 0 for all n, {Vsn_1}*==2 converges to V L Now observe that since sn = \ / 2 + \A«-i» the sequence

converges to v 2 + V L , and also that {.sn}~=i converges to L; thus, by the uniqueness of the limit of a sequence, L = \ / 2 + y/L. By eliminating the radicals, we see that L must be a root of the polynomial equation L4 - 4L2 — L -f 4 = 0. • It is left as an exercise to consider the sequence {(1 + ^)M}n=i • This sequence may be shown to be increasing and bounded, and hence convergent. The limit happens to be e, the base for the natural logarithm—a number that plays a central role in calculus. See Project 1.1 at the end of the chapter. 1.17 THEOREM Let £ be a set of real numbers. Then x0 is an accumulation point of E iff there is a sequence {xn}~=1 of members of E, each distinct from *0, such that {xn}~=1 converges to x0. Proof Let JC0 be an accumulation point of £. Then, for each positive integer n, there is a point xnE E such that 0 < \xn - x0\ < \. [The set (x0 — \y x0 + J) is a neighborhood of JC0, so it contains a member of E distinct from x0.] It remains to be shown that {xn}„=1 converges to x0. Choose e > 0. Then there is a positive integer N such that ^ < JV; hence, for w ^ iV, |xw - x0\ < £ ^ ^ ^ e. Thus, {*„}*»! converges to x0. Suppose now there is a sequence {xn}™=l of members of E, each distinct from x0, that converges to x0. Since every neighborhood of x0 contains all but a finite number of terms of the sequence, every neighborhood of x0 must contain at least one member of E that is distinct from x0. Hence, x0 is an accumulation point of E. We close this chapter by examining examples involving techniques that are of sufficient interest to be studied with care. • Example 1.7 Consider a real number 0 < b < 1 and form the sequence {bn}n=\* Observe that b"~l - bn - bn~l{\ -b) > 0; hence, the sequence is decreasing. In order to show that this sequence converges, it suffices to show that it is bounded from below, an easy task since bn > 0 for all natural numbers n. Although we know now that the sequence converges, we shall seek also to determine its limit. Now [bn)n=\ converges; call its limit L. And {b2n}™=u being a subsequence of {&n}~=i, also converges to L. On the other hand, {b2n}™=u which may be considered as the product of the sequence {bn}™=l with itself, converges to L2. By the uniqueness of the limit, L2 = L, and so L = 0 or L = 1. Clearly, L # 1 since the sequence was decreasing and

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1.4 Subsequences and Monotone Sequences 53

b < 1. Hence, {bn} ~= x converges to 0. This example is not intended to amaze the reader, but it illustrates a few techniques that are convenient to have at hand. • •

Example 1.8 Consider 0 < c < 1 and the sequence {^/c}™=l. For all n,

Vc-"\rC

= Vc(l - n("-\^) > 0,

since c < 1. Hence, the sequence is increasing, and clearly Vc < 1 for all n. Thus, the sequence converges; call the limit L. By Exercise 28, if {xn}*=1 converges to x0, xn ^ 0 for all n, then {V^}~=i converges to VxjJ. Therefore, the sequence { \ / ^ } r = i converges to VZ, but and {Vc}~=i is a subsequence of {Ve}~=i and hence converges to L. By the uniqueness of the limit of a sequence, VE = L; hence L = 0 or L = 1. But c > 0 and the sequence is increasing; therefore L # 0. In conclusion, {tfc)n=i converges to 1. (See Exercise 38 for the case c > 1.) • • Example 1.9 We shall now consider another sequence that is defined recursively—that is, the first few terms are given, and instructions are provided for computing the nth term in terms of some or all of the preceding terms. Let ax = 1, and n ^ 2, define an — v2a~i. We shall also use induction to attack this sequence. First, look at a few terms of the sequence,

ax = 1,

a2 = V2,

a3 = V i V l

It seems reasonable to expect the sequence to be increasing and, hence, convergent if it is bounded. It may not be clear atfirstglance what to try for an upper bound, so we shall discover this bound by a very useful device. (Even if you can guess the bound, play the game to see how the device works.) Since the sequence is increasing, if it converges, the limit will be the least upper bound of the sequence. In an attempt to discover a logical candidate for this limit, let us assume that the sequence converges. Suppose {an}n=i converges to L; then {aj}*=1 converges to L2, but a2n = 2tfw_1; hence, { 0. There is Ni such that for n ^ Nx \an - A\ < §, and there is iV2 such that for n S: JV~2 \an — B| < f. Use the triangle inequality to show that this implies that \A - B\ < e. Argue that A = £. 13. Let x be any positive real number, and define a sequence (a„}*=i by

fl

=

M + [2x] + • • • + [nx] =

where [JC] is the largest integer less than or equal to x. Prove that [an}^x converges to x/2.

12 CAUCHY SEQUENCES 14. Prove that every Cauchy sequence is bounded (Theorem 1.4). 15. Prove directly (do not use Theorem 1.8) that, if {a„}"=1 and [bn)n=\ are Cauchy, so is 16. Prove directly (do not use Theorem 1.9) that, if {a„)*=1 and {bn)n=\ are Cauchy, so is {tf,A}n=i- You will want to use Theorem 1.4. 17. Prove that the sequence < 18. 19. 20. 21. *22.

\

is Cauchy.

Give an example of a set with exactly two accumulation points. Give an example of a set with a countably infinite set of accumulation points. Give an example of a set that contains each of its accumulation points. Determine the accumulation points of the set {2n + | : n and k are positive integers}. Let S be a nonempty set of real numbers that is bounded from above (below) and let x - sup S (inf S). Prove that either x belongs to S or x is an accumulation point of S.

23. Let a0 and at be distinct real numbers. Define a„ = n ^ 2. Show that {an)^x that

for each positive integer

is a Cauchy sequence. You may want to use induction to show

( lX
0, then Vdu - Va = —~—^7=.) Van + Va 29. Prove that

Cr)

L(/i + *)*J»-.i converges to •$, where //* + k\

V * /

=

(a + *)!

«!*! '

30. Prove the following variation on Lemma 1.10. If {&„}".! converges to B =£ 0 and bn # 0 for all /?, then there is M > 0 such that |6 n | ^ M for all «. 31. Consider a sequence {an}~=1 and* for each n, define 2, 1 n+1 k \n + 1 5. Prove that the sequence

(HI

is increasing. 6. Items 3 and 5 show that the sequence

{(-01

is bounded and increasing; hence, it is convergent; call the limit e. Compute the 100th, 1000th, and 100,000th term of the sequence to approximate e. Compare your results with the number e that your calculator will display.

PROJECT 1.2 The purpose of this project is to establish the convergence of an iterative process for approximating square roots. The process is a familiar one—the divide-and-average process.

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Project 1.4 59 Define a sequence recursively as follows: ax = 2, and for « ^ 2 , define

fl =

- 5 («•-> + zr)

1. Prove that a„ ^ V2 for all w. This is a bit tricky. Use some algebra to write (an + ^ ) 2 - 8 as a square to establish the fact that (an + ^ ) 2 ^ 8. 2. Prove that the sequence is decreasing. 3. Prove that the sequence converges to V2. 4. Generalize this process tofinda sequence that converges to Va where a is any positive real number.

PROJECT 1.3 In Exercise 40, you proved that a particular sequence defined recursively was convergent. In this project, you will have an opportunity to generalize that result. Define a sequence by ax = a, and for n ^ 2, define a„ - Va + an-x. For what choices of a (positive, of course) will this sequence converge and to what limit? Prove your conjecture.

PROJECT 1.4

-.

In this project you will prove that the set of real numbers is not a countable set. The common usage is to say that such a set is uncountable. We begin with a few facts about equivalent sets of real numbers. Recall that sets A and B are equivalent if there is a 1-1 function from A onto B. Your strategy will be the following: You will show that [0,1] is equivalent to R and that [0,1] is uncountable. Then you can conclude that R is uncountable. First, you will show that [0,1] is equivalent to R. (You may want to refer to Project 0.1.) 1. Prove that R is equivalent to (0, 1), and (0,1) is equivalent to [0,1]. Conclude that R is equivalent to [0, 1]. Now your task is to prove that [0, 1] is uncountable. Your strategy will be to consider an arbitrary function / : / -» [0, 1] and prove that im / ¥* [0, 1]. Thus, there is no function from / onto [0, 1], and so [0, 1] is uncountable. Suppose that T is a function from / to [0, 1], 2. Show that there are sequences {an}"=1 and {6„}£=i such that [au bY] C [0,1], and for each nEJf [an+u bn+l] C [an, bn ] and T(n) g [am bn]. 3. Show that [an]%=i converges; call the limit A. 4. Prove that A G [an, bn] for each « 6 / , Conclude that A & im T. 5. Finish the proof that [0, 1] is uncountable.

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60 Chapter 1 Sequences

PROJECT 1.5 In this project, you will prove that every sequence has a monotone subsequence and then, without using the Bolzano-Weierstrass Theorem, prove that every Cauchy sequence converges. THEOREM Every sequence has a monotone subsequence. Proof 1. For a sequence {an}"=u prove that exactly one of the following statements is true: i. There is N such that no term of the sequence {aN+k}£=i is an upper bound of that sequence. ii. For each N, there is a term of the sequence {aN+k)k=i which is an upper bound for that sequence. 2. Prove that {an]™=x has a monotone subsequence as follows: i. If (i) in item 1 is true, construct an increasing sequence that is a subsequence of {an}™={. ii. If (ii) of item 1 is true, construct a decreasing sequence that is a subsequence of {an)n=x. 1.7 THEOREM Every Cauchy sequence is convergent. Proof

Use the theorem above.

PROJECT 1.6 Define a sequence recursively as follows: ax = 0, a2 = 1, and for n ^ 2, (*)fln+l =

j

*

1. View (*) as a difference equation and seek solutions of the form an - W1 for some w E R. You will find two numbers, wx and w2 that provide solutions to the difference equation. Now look for a and b such that an = aw" 4- bw\ gives the sequence as defined above. Does this look familiar? It should, especially if you have studied differential equations. 2. Show that the sequence converges and find the limit.

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Project 1.7 61

PROJECT 1.7 In this project, you will provide a proof, different from that in the text or Project 1.5, of the fact that every Cauchy sequence is convergent. In the process, you will give another proof that every bounded sequence has a convergent subsequence. 1. Let {a„}n=\ be a bounded sequence. For each n E / , define yn = $up{ak:k > n}. i. Prove that {yn} ~= i is decreasing and bounded from below, hence convergent. ii. Let A be the limit of the sequence \yn}™~i. Prove that \an}^ has a subsequence that converges to A. 2. Use item 1 to prove that every Cauchy sequence is convergent.

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Chapter

Limits of Functions

r

he notion of the existence of a limit of a function at a point underlies the study of calculus. When students fail to understand this idea, their study of calculus becomes a drudgery of juggling formulas. Before considering continuity, differentiation, and integration, one needs a thorough understanding of limits. The material on sequences presented in Chapter 1 will help in achieving this understanding. First we shall consider an example commonly found in calculus books. Let x2 - 1

/« = X-—r ~ 1 for x * 1, and let /(l) = 6. Wefindthat f(x) = x + 1 for x # 1 and /(l) = 6. To the unsophisticated, it may seem objectionable to define a function this way. Since / behaves so well elsewhere, it seems reasonable to define /(I) = 1 + 1 = 2. In this chapter we shall attempt to rearrange any prejudices of the reader concerning the behavior of such functions.

2.1 DEFINITION OF THE LIMIT OF A FUNCTION First of all, let us examine the intuitive idea, presented in elementary calculus, of a limit of a function. Consider a function / :D -> R where D C R and L is a real number.1 If x0 E R, then / has a limit L at x0 if, as x approaches x0, f(x) gets close to L. Of course, the values of JC to be considered must belong to the domain of the function. The precise formulation of the idea of getting close no longer poses any problem, since this has already been done in Chapter 1 while dealing with convergent sequences. To summarize, in order to formalize the definition of a limit of a function, it is necessary to define precisely what is meant by '*/(*) gets close to L as x approaches JC0." To set the stage ! Here and throughout the remainder of this book, R will be the set of all real numbers, and, unless specified otherwise, the domain of any functions considered will be subsets of R.

63

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64 Chapter 2 Limits of Functions

for this, it is necessary that there be points in D as close to x0 as one may wish. Finally, the limit should depend only on the behavior of the function near JC0, not on f(x0)', indeed, x0 need not even belong to D. DEFINITION Let / : D -> R with x0 an accumulation point of £>. Then / has a limit L at x0 iff for each e > 0 there is a 8 > 0 such that if 0 < \x — x0| < 8 and x E D, then \f(x) ~ L\ < e Observe that the definition is similar to that concerning a sequence converging to a real number L in that it does not specify that if L exists, it is unique. However, this is true, and the proof is left as Exercise 5. If / has a limit L at x0, we write L = l i m ^ f{x). It is incorrect for students to think they should memorize all definitions such as the preceding one. That this mistaken impression is widespread among students is evidenced by the fact that many are able to regurgitate the words of the definition, sometimes in the wrong order, without understanding the full meaning. A definition should be carefully examined, dissected, and fully assimilated. The reasons behind it and the full meaning of each part of the statement should be carefully considered. The ideas should become not only a part of one's memory, but a part of one's way of thinking— familiar friends and companions along the journey to enjoyment of mathematics. First, the insistence that xQ be an accumulation point of D, the domain of /, is necessary in order that / be defined at points near x0. Now, the positive number e is the desired degree of closeness chosen in advance. To fulfill the definition, one must find 8 > 0 (which will generally depend on e) such that if x E (jt0 — 8, x0 + 8), x E D, and x =£ x0 (the behavior of / at x0 should not affect the limit); then fix) E (L - e, L + e). This last statement is equivalent to insisting that the graph of / for x E DOixo - 8, Jt0 + 8), x ¥= JC0, lies in the strip {(JC, y): L - €< y < L 4- e}. (See Figure 2.1.) The hole in the graph emphasizes that the value of / at x0 is not considered when determining whether L is the limit of / at x0. m Example 2.1 Let us now reconsider the example mentioned at the beginning of the chapter. Let D = /?, and define / : D -> R by

for x ^ 1 and /(I) = 6. For x * 1, fix) =

= x + 1; hence, / is a linear x- 1 function, and the graph of / is a line with slope 1, except for x = 1. As mentioned above, the number /(l) has nothing to do with the existence of a limit at x = 1. Thus, one is led to believe (correctly, by the geometry of the plane) that as x approaches 1, fix) approaches the value necessary to fill in the gap in the line—namely, 1 + 1 = 2 . Let us prove that / has a limit L = 2 at x = 1. Choose € > 0. Consider the geometric interpretation of the idea of a limit. It is necessary to choose a neighborhood of 1 such that for x in this neighborhood with x # 1, the corresponding points on the graph of / lie in the strip { (JC, y): 2 - € < y < 2 + e}. Ignoring the point x = 1, the graph of / is a straight line

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2.1 Definition of the Limit of a Function 65

> x Figure 2.1

of slope 1. Thus, one is tempted to try €= 5 to obtain the neighborhood (1 - 8,1 + S) of x = 1. This result could be obtained by direct computation, but it is worthwhile to take this opportunity to emphasize the geometric aspect of the definition. If 0 < |JC - 11 < 8 = e, then | / « - 2| =

1

x-

1

= |(x + 1) - 2| = |JC — 1| < 8 = e.

and the conjecture is shown to be correct. See Figure 2.2.

•

In the next example, we exhibit a function that fails to have a limit at a point. In order to show that a function does not have a limit at a point, we must show that no number L can meet the requirements of the definition. The strategy will be to assume that the function has a limit and arrive at a contradiction, showing that the limit does not exist. Later we will have simpler methods of showing that a function fails to have a limit at some point. \x\ Define / :R\{0] -*i?by/(x) = —. For* < 0, /(*) = - 1 and x fovx > 0, f(x) = 1. This suggests that / does not have a limit at zero. We will proceed m Example2.2

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66 Chapter 2 Limits of Functions

> x

by assuming there is a number L that satisfies the requirements of the definition of the limit of / at zero and arrive at a contradiction. To the left of zero the values of the function are - 1 , and to the right of zero, the values are 1. So if we choose e = 1/4, there is no number L that differs from - 1 by less than 1/4 and that differs from 1 by less than 1/4. Assume L is a limit of / at zero. Choose e = 1/4. There is 8 > 0 such that if 0 < \x - 0| < 5, then \f(x) - L\ < 1/4. Choose p = -8/4 and q = 8/4. Then /(/>)= - l , / ( ? ) = l,andso 2 = |/(p) - f(q)\ ^ \f(p) - L\ + \L - f{q)\ < 1/4 + 1/4 = 1/2. This is a contradiction. The conclusion is that / does not have a limit at zero.

•

In the next example, we will show that the given function fails to have a limit at 0. This time the strategy will be to let L be any real number and exhibit an e > 0 for which no 5 > 0 can be found that meets the requirements of the definition. Thus, L (an arbitrary real number) cannot be a limit of / at 0; hence, / does not have a limit at 0. The differences in the strategies used in Examples 2.2 and 2.3 are somewhat subtle. To comprehend more fully how to show that a function does not have a limit at a point,

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2.1 Definition of the Limit of a Function 67

you might want to use the strategy of Example 2.2 on the function in Example 2.3 and vice versa. Not only will this aid in your understanding of limits of functions, but it will illuminate the differences between the two strategies. • Example 2.3 Define / : (0,1) -* R by f{x) = sin j , and consider the behavior of / at x = 0. Although zero is not in the domain of /, it is an accumulation point of the domain of /; hence, it is reasonable to inquire whether / has a limit at zero. In this example, it is instructive to look at the graph of this function in Figure 2.3 to gain insight into its behavior as x approaches zero. Since / (l/nir) = 0 for each positive integer n and / (2/nrr) is 4-1 or - 1 for n an odd positive integer, we see that the function oscillates more and more wildly as x becomes close to zero. Consequently, it is reasonable to conjecture that / does not have a limit at zero. Let L be any real number and choose e = 1/3. To show that L is not a limit of / at zero, it suffices to show that for any S > 0, there is x E (0, 1) such that 0 < |* - 0| < S and \f(x) - L| > e. By the choice of e, either 1 (£ (L - e, L + e) or 0 £ (L - €, L + €). Suppose the latter is the case. Then, given S > 0, there is a positive integer n such that l/rnr < 8, and, as observed above, f(l/mf) = 0 $• (L - €, L + e); hence, \f{\lmr) - L\ ^ c. The case for l £ (L - e, L + e) can be handled similarly. Thus, / does not have a limit at zero. •

Figure 2.3

• Example 2.4 Define / : (0, l) -» R by f(x) = I/JC. Now zero is an accumulation point of (0, I); hence, we may inquire as to the existence of the limit of / at zero. Again, from a sketch of part of the graph of this function, we can see that the function *'blows up" at zero. This is not the sort of behavior one expects a function to have if it is to have a limit at zero, so let us prove that / does not have a limit at zero. Let L be any real number, and choose e > 0 such that L + e > 0. Now, if 0 < x
e. Thus, it is impossible to find a 8 > 0

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6$ Chapter 2 Limits of Functions y

*>-i

\

1

Figure 2.4

to fulfill the requirements of the definition; that is, L is not a limit of / at zero. Since L is any real number, / does not have a limit at zero. (See Figure 2.4.) • • Example 2.5 Let us now consider a more sophisticated example. Define / : [0, 1] -> R by f(x) = 0 if x is irrational; and if x is rational, set f(x) = \lq where x = plq with p and q nonnegative integers that are relatively prime. Thus, we have

«»"•'(*)-}•'(H'(TH ' and so on. We shall seek to determine those points at which / has a limit and those at which / does not have a limit. Suppose x0 E [0, 1], and let us examine the behavior of / near x0. Since there are irrational points in every neighborhood of x0 that are distinctfromx0, it is clear that / takes the value 0 in every neighborhood of x0 infinitely often. Thus, if there is to be a limit of / at x0, it is clear that the limit must be zero. It remains to decide if it is reasonable to expect that / has zero as a limit at xQ. For x E [0, 1], f(x) is small (close to zero) only if x is irrational or if x = plq and p and q are relatively prime with q large. However, for afixedpositive integer q, there are but afinitenumber of points in [0, 1] of the form plq; in fact, for a fixed positive integer q0i there are only a finite number of points in [0,1] of the form plq, where q^qo with p and q positive integers. Thus, f(x) >. llq0 at only afinitenumber of points. In view of these observations, the reader should suspect that indeed / has a limit at JC0 and the limit is zero. Let us now give a proof of this fact. Choose € > 0. There is a positive integer q0 such that l/q0 < €. There are at most a finite number of rational points in [0, 1] of the form plq where p and q are positive integers with q < q0, say ru . . . , rn. We may assume that JC0 is deleted from this list if

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22 Limits of Functions and Sequences 69

it should happen to be of this form. Now, to guarantee that f(x) is small, it is sufficient to avoid these points. Thus, let S = min{ \XQ - n\: / = 1 , . . . , n], and observe that 8 > 0. Now if 0 < \x - x0\ < S and x G [0, 1], then x is either irrational, in which case f(x) = 0, or x = p/q where p and q are relatively prime with q > q0, in which case f(x) = l/q. In either case,

\f(X) - 0\ < l/Wl = £ - < - < €. Thus, / has a limit at each x G [0, 1], and that limit is zero.

•

A word of philosophy seems appropriate at this point. The emphasis on rigor, both in definitions and proofs, is intentional because this is the only way mathematics can be communicated intelligently. However, one's feelings and intuition about these matters should not be discarded or considered unimportant. On the contrary, intuition is very important and must be cultivated, but it must also be accompanied by the realization that we must provide rigorous proofs for those facts that our intuition tells us to be true. Finally, the intuition must constantly be readjusted because we are occasionally led astray by our feelings. In Example 2.3, the oscillatory nature of the function, as evidenced by a sketch of the graph, led our intuition to tell us that the function does not have a limit at zero. The behavior of the function in the preceding example will probably elude the intuition of the novice. However, the facts are now before us and should be considered as a new experience leading to greater mathematical maturity.

2.2 LIMITS OF FUNCTIONS AND SEQUENCES As mentioned in Chapter 1, a strong relationship exists between limits of sequences and limits of functions. Now is clearly the time to seek out these relationships and exploit them to the fullest. Suppose / :D -» R with x0 an accumulation point of D, and suppose / has a limit L at x0. Consider a sequence {xn)n=i converging to x0 with xn E D, x0 # xm for each positive integer n. Since / has a limit L at x0, as the terms of the sequence get close to x0, the corresponding values of / must get close to L; in fact, we must have {/(*J}£L, converging to L. Fortunately, the converse of this theorem is also true. We summarize thse results in the following theorem. 2.1 THEOREM Let / : D -> R with x0 an accumulation point of D. Then / has a limit at x0 iff for each sequence {xw}~=1 converging to x0 with xn G D and xn # x0 for all n, the sequence {f(xn)}%=l converges. Before proving this theorem, reflect on one aspect of its content. Consider two sequences U,}"., and {yX=i such that xn, yn G D, xn # x0, yn ± x0 for n = 1, 2 , . . . , and such that both {x„}^l and {yX = 1 converge to xQ. If we assume that the latter

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70 Chapter 2 Limits of Functions

condition of the theorem holds, then {/0e„)}~=i and {/()>„) }"«i both converge to, say, Lu and L2 respectively. (Note that the condition does not assert that Lx - L2.) Form a new sequence {z„}~=1 where z2n = xn and z2n-\ = yn- This sequence consists of members of D distinct from *0, and it converges to JC0; hence {/(zj}~=i converges. In particular, {/(xn)}"=1 and {f(yn))n=i are subsequences of the convergent sequence {f(,Zn))n=i and hence have the same limit; that is, Lx = L2. Now, if for every sequence {*„}£=! of members of D distinct from x0 and converging to x0, the sequence {/(*/i)}*=i converges, then all such sequences have a common limit by use of the preceding observations. This limit should be the limit of the function, a fact that remains to be seen. Proof Suppose / has a limit L at x0. Let {JC„}~=1 be a sequence of members of D distinct from x0 but converging to JC0, and consider the sequence {/0t„)}~=1. Choose e > 0. There is 8 > 0 such that if 0 < \x - x0\ < 8 with x E D, then \f(x) - L\ < €. Since {*„}~=1 converges to x0, there is N such that for n ^ N, \xn - x0\ < 8. Now for n ^ N, 0 < \xn - x0\ < 8 and xn E D\ hence \f(xn) - L\ < €. Thus, {/CO}*=1 converges; indeed, it converges to L. Suppose now that the latter condition is satisfied and so by the remarks following the theorem, all the sequences {f(xn) }*= { have a common limit, called (with great originality) L. Suppose that L is not a limit of / at x0. (We do not assume anything concerning the existence of the limit of / at *0; we assume only that L is not a limit of / at JC0.) Thus, there is € > 0 such that for every 8 > 0, there is x E D, with 0 < \x - x0\ < 8 and such that \f(x) - L\ > e. In particular, for each positive integer ny there is xn E D with 0 < \xn - xQ\ < \ such that \f(xn) — L\ > e. The sequence {JC„}~=I converges to x0 and is a sequence of members of D distinct from x0; hence, {/Oc„)}~=i converges to L, contrary to the fact that \f(xn) - L\ ^ e > 0 for all n. Thus, L must be the limit of / at x0. To summarize, / has limit L at x0 iff, for each sequence {xw}*=1 of members of D distinct from x0 converging to JC0, the sequence {f(x„)}~=l converges. If the second condition is satisfied, then all the given sequences {/C*„)}£=i have a common limit, which is L, the limit of the function / . An example should serve to reveal the usefulness of this equivalence and the related observation that follows it. Let / : (0, 1) —> R satisfy the following condition: there is K > 0 such that, for all x, y E (0, 1), \f(x) - f(y)\ l, x0 + (~l)M^ < *0, and thus

/ U + (-irH =*o- l; and for n even, JC0 + (—l)"^ > *o» ^ d thus

/ U + (~ir~) =*o.

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72 Chapter 2 Limits of Functions

2 U

-O

Kx)=[x]

1 U -•

x

-4

-O

h-2

Figure 2.5 Thus, {/(x0 + (""l)n«)l"=i does not converge, and so / does not have a limit of JCQ if x0 is an integer. • Consider the examples /(JC) = [JC], g(jc) = sin £, and h(x) = J. Each failed to have a limit at zero for basically different reasons. The function /(JC) = [x] "jumped" at zero, taking the value - 1 to the left of zero and the value 0 to the right of zero; g oscillated too badly at zero to have a limit there; h was unbounded near zero and so could not have a limit there. In a sense, these three examples describe the sort of behavior one may suspect of a function that does not have a limit at a point. We shall comment more on this in Section 2.4.

2.3 ALGEBRA OF LIMITS Following the pattern set in Chapter 1, we shall now determine the relationships between limits of functions and the algebraic operations. If / : D —> R and g:D -> R, define / + g:D -» R by (/ + g)(x) = /(JC) 4- g(jc) for all JC G D. In a similar fashion, define fg : D -> R by (fg)(x) = f(x)g(x). If g(x) * o for JC G D, we may define lg : D -» R by

for all JC G D. Now read Theorems 1.8, 1.9, and 1.11, and attempt to formulate corresponding theorems for limits of functions. In each case, the proof can be obtained by use of Theorem 2.1 and the appropriate result from Chapter 1.

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23 Algebra of Limits 73 2.4 THEOREM Suppose f,g:D-» R with x0 an accumulation point of D, and further suppose that / and g have limits at x0. Then: i. / + g has a limit at *0, and limCf + g)(x) = lim f(x) + lim g(x). X-*XQ

X-*XQ

*-*XQ

ii. fg has a limit at x0 and lim(fg)(x) = [lim / and lim^^0 g(x) # 0, then \ has a limit at x0 and lim

/.\

/to

lim (M to ~* :^ \ g / lim g(x) 0

The proofs of (i) and (iii) will be accomplished by use of sequences, whereas the proof of (ii) will be direct. Proof (i) Let {xw}~=i be any sequence of points in D converging to x0 with x„ i= x0 for all w. It suffices to show that {(/ 4- £)(*„) }£=i converges to lim /(x) + lim g(x). X-*XQ

X-+XQ

By assumption, / and g have limits at x0\ hence, {/(x„)}««i converges to lim*-^ f{x) and {g(x„)}?.i converges to lim*^ g(x). By Theorem 1.8, {(/ + g)(*„)}«=i = if(xn) + *fc)ir-i converges to lim*^, f(x) + lim^_^0 g(x). This concludes the proof of (i). (ii) Given e > 0, we must find an appropriate S > 0. (Before beginning the proof, see the discussion preceding Theorem 1.9.) As the plan for finding die 5 unfolds, observe the parallel between this search and that for N in Theorem 1.9. Choose €> 0. Let A = lim /(JC)

and

X—*XQ

B = lim g(x). X-*XQ

We wish to find 8 > 0 such that if 0 < \x - x0\ < 8 and x E Df then \f(x)g(x) - AB\ < e. By Theorem 2.3, there is 8X > 0 and a real number M > 0 such that for 0 < |JC - x§\ < 8U x 6 D , w e have |/(x)| ^ M. Let £ +M

>0.

There is 82 > 0 such that if 0 < |JC - x0\ < 82, x E D, then |/(x) - A\ < e'; and there is S3 > 0 such that for 0 < \x - XQ\ < S3, x E D, then \g(x) - J?| < e\ Let 8 = minf^, ^2, 83).

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74 Chapter 2 Limits of Functions

Now, if 0 < \x - XQ\ < S, x E D, then \(fg)(x) -AB\ = \f(x)g(x) - AB\ < |/(*)*(x) - /(JC)B| + |/(*)£ - AB| = \f(x)\\g(x) - B| + |B||/(JC) - A| < M€' + |B|€' = e. (iii) Suppose {*„};£= i is a sequence of members of D, distinct from x0, converging to *0. Now {/(*„)}£., converges to lim^^0 f(x) and {g(xn)}%=1 converges to limx_^0 g(x). Since g(x) ^ 0 for all x E D, g(xn) =£ 0 for all n. By assumption, lim*..^ g(x) =£ 0. Hence,

{©«}>{££

converges to

lim f(x) X-+XQ

lim g(x) X-*XQ

By Theorem 2.1, J has the indicated limit at JC0. The adventuresome reader may wish to attempt a direct proof of (iii). A lemma similar to Lemma 1.10 is in order, and then the proof may be patterned after that of Theorem 1.11. See Exercise 21. Good luck! 2.5 THEOREM Suppose / : D —> R and g : D -» R, x0 is an accumulation point of £>, and / and g have limits at x0. If f(x) ^ g(x) for all x E D, then lim f(x) < lim g(x). X-+XQ

X-^XQ

Proof The proof is left as Exercise 20. Let us now look at an example suggested by the sequence [\ sin if }~=1. Consider the function / : (0, 1) -» R, defined by f(x) = x sin £. It has been shown that sin j fails to have a limit at zero, so we may not use Theorem 2.4 (ii). However, sin£ is bounded above by 1 and below by - 1 . Now it is clear that . l

I/Ml = x sin x- * W : hence, / has a limit at zero; in fact, lim^o /(*) = 0. (We merely choose 8 = e) A theorem reminiscent of Theorem 1.13 is in order. 2.6 THEOREM Let / : D -» R and g : D -» # and JC0 be an accumulation point of D. If / is bounded in a neighborhood of x0 and g has limit zero at x0, then fg has a limit at x0 and lim*.^ /g(x) = 0.

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2.3 Algebra of Limits 75 Proof Choose e > 0. There is a 5, > 0 and M > 0 such that if JC E D and \x - x0\ < Su then |/(JC)| ^ M. Let e' = c/M. There is ^ > 0 such that if x E D and 0 < \x - x0\ < S2l then \g(x) - lim g(x)\ = \g(x)\ < €'. Choose 5 = minfSj, 52}. Then if 0 < |JC - x0\ < 8 with x E D, a computation yields \(fg)(x)\ = |/(*)*(*)| = \m\\g(x)\

* Me' = e.

Hence, fg has a limit at x0, and limJC_>JCo(/g)(jc) = 0. The proper use of the preceding theorems will allow us to handle a rather large class of functions. Suppose / : R —»R is defined by f(x) = x for each x E R. If x0 E R, then for each sequence {JC„}^=I converging to x0 with xn E R\ {XQ}, {/H.

Suppose {JCI, . . . , xk) CJn with xx < x2 < • • * < xk. Now if / has a jump greater than 1/n at JCI, . . . , xkf then it seems reasonable to suspect that k can't be too large; in fact, k should be less than or equal to n[f(fi) - /(a)]. Thus, /„ should befinite,and the set of points where / fails to have a limit should be U~=1/„, a countable union of finite sets, hence a set that is countable. This is the content of the next theorem; we have outlined the proof in a vague way and must now fill in the details. First we prove a lemma. 2.7 LEMMA Let / : [a, /3] -~» R be increasing. Let U(x) = M{f(y) :x 0. There is 8 > 0 such that if 0 < \x - x0\ < 8 with x E [a, 0], then \f(x) - A\ < e. Since x0 E (a, 0), there are x,y E [a, 0] such that x0 - 8 < x < x0 < y < x0 + 8\ so, by the definition of L and U and the fact that / is increasing, A - e < f(x) *s L(x0) < /(xo) =s £/(*0) ^ /(J) < A + e. Therefore, £/(x0) - £(*o) < 2e for each e > 0; hence, £/(*) = L(^0); hence ^(^o) = f&o) = t/(xb). Moreover, A - € < (7(JC0) < A + e for all € > 0; hence, £/(JC0) = A. Thus, lim /(*) = U(x0) = L(x0) = /(x0). X-*XQ

Suppose now that f/(jc0) = L(x0). As observed above, L(x0) ^ /(JC0) ^ t/(jc0); hence U(x0) = f(x0) = L(x0). It remains to be shown that / has a limit at x0 and

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78 Chapter 2 Limits of Functions

that lim,.^ f(x) = /(%)• Choose e > 0. Now L(x0) — e is not an upper bound for if(y): y < xo)> and U(x0) + €is not a lower bound for {/(y): x0 < v}; hence, there are real numbers yx and v2 such* that a ^ yi < x0 < y2 ^ 0 and such that L(*0) - €< /(y,) and /(y2) < l/(x0) + e. Let 8 = min{x0 - yu v2 ~ *o}- Now if 0 < \x — x0\ < S, then yx < x < y2; hence, f(xo) - €= L(x0) - €< f(yx) 0. Let Dn = \x:U(x) - L ( x ) > - | . It is clear that D = U~=iDw. The proof will be complete if we can show that each Dn isfinite.Suppose {xu . . . , xr) CDn with a < JC, < x2 < • • • < xr < 0. Choose zu . . . , zr+1 such that a < z, < JC!, JC,- < z,-+i < ;cl+1 for i = 1, 2 , . . . , r — 1, and xr < zr+1 < 0. Now for each j, /(z,) ^ Lfo) and [/(*/) ^ /(z /+ i); hence / ( z m ) - /(z,) s> £/(*,) - LOO > - . n Now

m - /(«) = /08) - f(zr+l) + X [/ R by f{x) = - . Prove that / has a limit at - 2 , andfindit. 2x2 -4- 3JC — 2 2. Define / : (-2, 0) -» R by /(JC) = ——. Prove that / has a limit at - 2 , and find it. 3. Give an example of a function / : (0,1) -» R that has a limit at every point of (0,1) except |. Use the definition of limit of a function to justify the example. 4. Give an example of a function f :R —> R that is bounded and has a limit at every point except - 2 . Use the definition to justify the example. *5. Suppose f :D -+ R with XQ an accumulation point of D. Assume Lx andL2 are limits of / at xQ. Prove Lx = L2. (Use only the definition; in later theorems, this uniqueness is assumed.) 6. Define / : (0, 1) -* R by f(x) = cos ±. Does / have a limit at 0? Justify. 7. Define / : (0,1) -* R by f(x) = x cos J. Does / have a limit at 0? Justify. x3 — JC2 + JC — 1 8. Define / : (0,1) -> /? by /(JC) = .Prove that / has a limit at 1. x— 1 JC + 1

9. Define / : (-1,1) -» # by /(*) = 2 _ . Does / have a limit at 1? Justify. 2.2 LIMITS OF FUNCTIONS AND SEQUENCES 10. Consider / : (0, 2) -> # defined by /(JC) = **. Assume that / has a limit at 0 andfindthat limit. (Hint: Choose a sequence {*„}"-,! converging to 0 such that the limit of the sequence {/(*n)}«=i is easy to determine.) •11. Suppose /, g, and h:D -> R where x0 is an accumulation point of D, f(x) ^ g(x) ^ /*(*) for all JC E D, and / and h have limits at JC0 with lim*.^ /(JC) = lim*.^ h(x). Prove that g has a limit at JC0 and lim X-+XQ

/(JC)

= lim g(x) = lim h(x). X-*XQ

X-*XQ

*12. Suppose / : D -> R has a limit at JC0. Prove that | / | : D ~> R has a limit at x0 and that linWo l/Wl = llim*-**o /Wl13. Define f :R->Rby f(x) = JC - [JC]. (See Example 2.6 for the definition of [JC].) Determine those points at which / has a limit, and justify your conclusions. 14. Define /:/?->/? as follows: /(JC) = 8JC if JC is a rational number. /(JC) = 2JC2 + 8 if JC is an irrational number. Use sequences to guess at which points / has a limit, then use c's and S's to justify your conclusions.

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80 Chapter 2 Limits of Functions 15. Let f :D -> R with xQ as an accumulation point of D. Prove that / has a limit at x0 if for each e > 0 , there is a neighborhood Q of x0 such that, for any x, y G Q D Dt x # JC0, y & x0, we have \f(x) - f(y)\ < e. 2.3 ALGEBRA OF LIMITS 16. Define / : (0, 1) -> R by /(JC) = limit. 17. Define f:R-^Ras

x3 + 6x2 + x ; . Prove that / has a limit at 0 andfindthat x — ox

follows:

/(JC) — x — [JC] if |X1 is even. /(*) = JC - [JC 4- 1] if [JC] is odd. Determine those points where / has a limit, and justify your conclusions. 18. Define g : (0, 1) -» /? by g(jc) =

vm -1

V ^ J C -

3

. Prove that g has a limit at 0 andfindit.

19. Define / : (0, 1) -» R by /(JC) = . Prove that / has a limit at 0 andfindit. *20. Prove Theorem 2.5. 21. Suppose g:D -> R with x0 an accumulation point of D and g(x) # 0 for all x G D. Further assume that g has a limit at x0 and lim^^^ g(x) ^ 0. State and prove a theorem similar to Lemma 1.10 for such a function. 22. Show by example that, even though / and g fail to have limits at x0, it is possible for / + g to have a limit at x0. Give similar examples for fg and |. 2.4 23. 24. 25.

LIMITS OF MONOTONE FUNCTIONS State and prove a lemma similar to Lemma 2.7 for decreasing functions. Let / : [a, h] —> R be monotone. Prove that / has a limit both at a and at b. Suppose / : [a, b]-*R and define g : [a, b] —• /? as follows: *(*) = sup{/(0:a < * < * } . Prove that g has a limit at xQ if / has a limit at JC0 and lim^^ f(t) = f(x0).

MISCELLANEOUS 26. Assume that f:R-+R is such that /(JC + y) = /(jc)/(y) for all x, y G R. If / has a limit at zero, prove that / has a limit at every point and either limx_0 /(•*) = 1 or f(x) = 0 for all xER. 27. Suppose / : D -» R, g : £ - » / ? , JC0 is an accumulation point of D 0 E, and there is e > 0 such that D fl [JC0 ~ e, JC0 + e] = E D [x0 - e, JC0 + e]. If f(x) = g(x) for all JC G D H £ (1 [JC0 - €, jc0 + e], prove that / has a limit at x0 iff g has a limit at x0.

PROJECT 2.1 The purpose of this project is to ascertain under what conditions an additive function has a limit at each point in R. We say that / : R -* R is additive if for all x, y E #, /(* + y) = /(JC) + /(y). In what follows, / is an additive function.

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Project 22

81

1. Show that for each positive integer n and each real number x, f(nx) = nf(x). 2. Suppose / is such that there are M > 0 and a > 0 such that if x E [-#, a], then \f(x)\ ^ M. Choose € > 0. There is a positive integer Af such that M/N < €. Show that if \x - y\ < a/N, then |/(JC) - f(y)\ < €. 3. Prove that if there are M > 0 and a > 0 such that if x E [-a, a], then \f(x)\ ^ M, then / has a limit at each x in R and l i n w /(f) = /(*). 4. Prove that if / has a limit at each x E R, then there are M > 0 and a > 0 such that if x E [-a, a], then |/(x)| < M. You have now proven that if / : R -» R is additive, then / has a limit at each point in R iff there are M > 0 and a > 0 such that if x E [—a, #], then \f(x)\ ^ M. In addition, if the condition is satisfied, limf_^. /(f) = /(JC).

PROJECT 2.2 In Exercise 39 of Chapter 1, you proved what may be called the "shuffle" theorem. In that exercise, you formed a new sequence by putting two sequences together in a special way and then examined the new sequence for convergence. The theorem you will be asked to prove in this project will be similar except that you will deal with functions and limits of functions rather than sequences. You might want to look at Exercise 14 in this chapter before reading further. 1. Prove the following theorem. THEOREM Suppose / : A -» R and g : B -> R are such that A n B = 0 , x0 is an accumulation point of A and also an accumulation point of B. Define h :A U B -> R by h(x) = f(x) if x E A and h(x) = g(x) if x E B. Prove that h has a limit at x0 if and only if / and g each have a limit at JC0 and l i m ^ /(x) = l i n v ^ g(x). You probably recall reading about one-sided limits in your calculus book. This theorem allows us to examine the relationship between one-sided limits and two-sided limits. We need a definition to set the stage. DEFINITION Let / :D —» R, and suppose x0 is an accumulation point of {x: x E D, x > x0]. Then / has a right-hand limit L at x0 iff for each e > 0, there is 8 > 0 such that if x0 < x < x0 + 8, x E D, then |/(JC) - L| < €. 2. State carefully the definition of left-hand limit for a function. 3. Give an example of a function that has a left-hand limit and a right-hand limit at a point but does not have a limit at that point. 4. Give an example of a function that has a left-hand limit and a right-hand limit at a point and also has a limit at that point. 5. State and prove a theorem that relates the left-hand limit and right-hand limit of a function and the limit of a function.

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82 Chapter 2 Limits of Functions

PROJECT 2.3 When examining a function to determine if it has a limit, it is sometimes convenient to make a change of variable. The theorem you will be asked to prove is designed to assist in using the change of variable in limits of functions. 1. Prove the following theorem. THEOREM Suppose / : A -» Rf g : B -* A. Suppose that a is an accumulation point of A, b is an accumulation point of B, i. lim,^* g(t) = a, ii. there is a neighborhood Q of b such that for t E Q n B, g(t) # a, and iii. / has a limit at a. Then / ° g has a limit at b and lim*.^ /(JC) = lim,_^ f(g(t)). 2. Consider f: # \ {0} -* Rby f(x) =

si n JC

. A standard argument in calculus is x to show that / has a limit 1 at zero. You may use this limit. Use the theorem you just proved to show that the function h:R\{ir/2] -» R defined by h(x) =

COS JC

2~X

has a limit at f, andfindthat limit.

PROJECT 2.4 Let / be a real-valued function defined on {x: x > a) for some real number a. In this project, you will examine the notion of having a limit "at infinity." DEFINITION If / is defined on {x: x > a} for some real number a, then / has a limit L at infinity if, for each e > 0, there is M such that for x > M and x > a, |/(x) - L\ < €.

1. Define f(x) = 2. Define f(x) =

Vx Vx

for x > 0. Prove that / has a limit at infinity, andfindit. 7= for x > 1. Prove that / has a limit at infinity, and find

1 — VJ C it.

3. If / is defined on {x:x> a}, define g(x) = /(£) for 0 < JC < £. Prove that / has a limit at infinity if and only if g has a limit at zero. 4. Define what it means for a function to have a limit at minus infinity. Illustrate with an example. 5. State and prove a theorem similar to that in item 3 for a function with a limit at minus infinity.

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Chapter

Continuity

n the discussion of the limit of a function it was emphasized that the value of / at x0 has no bearing on the question of the existence of the limit of a function / / at point x0. The concept of continuity of a function at a point brings the value of / at x0 back into the picture. In very imprecise language, a function / will be continuous at a point x0 if f(x) lies close to /(JC0) whenever x is sufficiently close to x0. Let us examine this notion in a familiar setting. Consider the function f(x) = Vx for x > 0. Because /(2) is an irrational number, it cannot be expressed by any of the conventional means as a decimal. Nevertheless, for purposes of computation, we can attempt to obtain rational numbers fairly close to V2. For example, (1.4)2 = 1.96, (1.41)2 = 1.9881, (1.414)2 = 1.999396, and so on. In essence, we are observing that /(1.96) = 1.4, /(1.9881) = 1.41, and /(1.999396) = 1.414, and we assume that 1.414 is close to V2 because 1.414 is the value of / at a point close to 2. Thus, the assumption is that f(x) = VJC is continuous at 2, a fact to be proven later. As with all new ideas, a precise definition is needed.

3.1 CONTINUITY OF A FUNCTION AT A POINT DEFINITION Suppose E C R and / : £ - » # . If x0 6 E, then / is continuous at x0 iff for each e > 0, there is a 8 > 0 such that if |JC

- x0\ < 5, x E £,

then

l/to " /C*o)| < e If / is continuous at x for every x E E, then we say / is continuous. Compare this definition with the definition in the previous chapter concerning the limit of a function at a point x0. First, for continuity at x0, the number JC0 must belong to Ef but it need not be an accumulation point of E. Indeed, if / : E -» R with x0E. E 83

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84 Chapter 3 Continuity

x0-§

x0

XQ + 8

>x

Figure 3.1

and x0 not an accumulation point of E, then there is 8 > 0 such that and x E E, then x = x0; hence

if\x-x0\ 0. In other words, if x0 is not aji accumulation point of E and Jt0 E £, then / is continuous at x0 by default. Thus, the only interesting case is when x0 is an accumulation point of £*. In this case, the definition requires that / has a limit at x0 and that that limit be /(JC0). This fact is a part of the next theorem, but before stating and proving that theorem, compare Figure 3.1 with Figure 2.1 of Chapter 2. Thus, / is continuous at x0 if and only if for each €> 0 there is a 8 > 0 such that the graph of / for x0 - 8 < x < x0 + 8f x E E, lies in the strip [(x, y): /(JC0) - € < y < f(x0\ + €}. 3.1 THEOREM Let f:E-*R withx0 ' E and x0 an accumulation point of E. Then (i) through (iii) are equivalent: i. / is continuous at xQ. ii. / has a limit at x0 and lim^^ f(x) = /(%)• iii. For every sequence {JCW}*=1 converging to x0 with xn E E for each n, {/(*«)}«=! converges to f(x0).

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3.1 Continuity of a Function at a Point 85 This theorem is the first of its type—that is, one that asserts that two or more statements are equivalent. Our method of proof will be to show that (iii) implies (ii), then show that (ii) implies (i), andfinallyshow that (i) implies (iii). This completes the cycle showing that (i), (ii), and (iii) are equivalent—that is, that each implies and is implied by the other. Also, if x0 is not an accumulation point of E with x0 E E, then (i) is always true, as is (iii). Exercise 4 asks you to prove that (iii) is true in this case. Proof Assume (iii) holds. In particular, if {*w}~=1 converges to x0 with xn # x0 and xnE E for all n, then {f(xn) J*=1 converges to f(x0). Hence, by Theorem 2.1, / has a limit at x0 and lim.^ 0 f(x) = f(x0). Thus, (iii) implies (ii). Assume (ii) holds, and choose e > 0. Since (ii) holds, there is 8 > 0 such that if0 0 such that for \x — x0\ < 8 and x E Ef | / t o - /(JC0)I < e. Since {*„}"«! converges to x0, there is N such that for n > N, \xn - x0\ < 8. Thus, for n > N, \f(xn) - f(x0)\ < e. This shows that {/toi)}«=i converges to f(x0). (Does this paragraph look familiar? It should. Compare it with the proof of Theorem 2.1.) It is worthwhile to reflect on this theorem and consider some of its less obvious uses. First, it can be used to show that a given function / is not continuous at a point x0 by exhibiting a sequence {xn}"=l converging to x0 where the sequence {f(xn)}%=i does not converge to f(x0). From a slightly different point of view, this theorem can be used to show that a certain sequence is convergent. For example, assume that the function / t o = ex is continuous on/?. Thus, in particular, if the sequence {jcn}*=1 converges to JC0, then the continuity of / t o = ex and Theorem 3.1 guarantee that [ex*)n**\ converges to ex°. We shall call attention to this aspect of Theorem 3.1 later. We pause momentarily to reconsider some of the results presented in Chapter 2. Recall the function / :R —» R defined by x2 - 1

/(*) = X-—f ~ 1

for x ¥= 1 and /(I) = 6. It has been shown that / has a limit at 1 and linr^ f(x) = 2. Unfortunately, lim^! / t o = 2 =£ 6 = /(I); hence, / is not continuous at 1. In this case, / fails to be continuous at 1 because /(l) does not happen to be the limit of/ at 1. Now consider the function g(x) = sin £ for 0 < x < 1, and g(0) = 38. We have seen that g fails to have a limit at zero, so g is not continuous at zero, regardless of what we try for g(0). The function h{x) = ~ for 0 < x < 1 behaves similarly in the

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86 Chapter 3 Continuity

sense that h doesn't have a limit at zero, so it is futile to attempt to define h(0) such that h will be continuous at zero. Recall the function / : [0, 1] -» R defined such that fix) = 0 if x is irrational and fiplq) = \lq if p and q are relatively prime nonnegative integers. It was shown in Chapter 2 that / has a limit at each point of [0,1] and that its limit is zero at each point. Thus, / is continuous at x E [0, 1] iff f(x) = 0 (that is, iff x is irrational).

3.2 ALGEBRA OF CONTINUOUS FUNCTIONS The next theorem is a natural follow-up to Theorem 2.4. 3.2 THEOREM Then

Suppose / : D -* R and g : D -> R are continuous at x0 E D.

i. / + g is continuous at x0. ii. fg is continuous at A*0. iii. If gix0) # 0, fig is continuous at x0. (A word of advice is necessary concerning (iii). The function fig is defined only where g{x) =£ 0; hence, when speaking of the function fig, we assume that the domain consists of those points where g(x) # 0. Lemma 3.3 will shed some light on this situation.) Proof If JC0 is not an accumulation point of D, then f + g, fg, and fig are continuous at x0 by default. So, for the rest of this proof, we shall assume x0 is an accumulation point of D. We shall use different methods of proof for items (i) and (ii) to illustrate the use of Theorem 3.1. A direct proof will be given for item (iii). Suppose / and g are continuous at JC0 E D. Let {xn} be any sequence of points in D that converges to x0. Then, by Theorem 3.1, {f(xn)} converges to f(x0) and U(*„)},?=i converges to g(x0); hence {/(*„) + *(*,,) }»-i converges to /(JC0) + g(x0). Thus,

{(/ + g)(xn)}n=i = if(xn) + *(*„)};r-i

converges to f(x0) + g(xQ) = (f + g)(x0), and hence, by Theorem 3.1, / 4- g is continuous at x0. This concludes the proof of (i). Suppose / and g are continuous at x0 E D. Then by Theorem 3.1, / and g each has a limit at JC0 (remember, we assumed xQ an accumulation point of D) and lim f(x) = f(x0)

and

X—>XQ

lim g(x) = g(x0). X-*XQ

By Theorem 2.4, fg has a limit at x0 and

limCfeX*) =

lim f(x)

x-*x0

lim g(x)

= f(xo)gix0)

Thus, by Theorem 3.1, fg is continuous at JC0.

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3.2 Algebra of Continuous Functions 87

We have purposely elected to give a direct proof of (iii) because it involves some important facts not yet brought to light. We hope the reader has attempted a direct proof of Theorem 2.4(iii) and discovered the need for a lemma similar to the following. The proof of Theorem 3.2(iii) follows this lemma. 33 LEMMA Let g : D -> R be continuous at x0 E D with g(x0) i= 0. Then there are 8 > 0 and a > 0 such that if \x - x0\ < 8 and x E D, then \g(x)\ ^ a. (Note the similarity between this lemma and Lemma 1.10, the one needed for the theorem concerning the convergence of the quotient of two convergent sequences.) Proof

Choose

a—

lg(*6)l > 0 .

There is 8 > 0 such that for \x - x0\ < 8, x E D, we have \g(x) - g(x0)\ < a. Thus, \g(x)\ > \g(x0)\ - \g(x) - g(x 0 )| > |*(*6)| - *

\g(x0)\

a.

Proof of 32 continued. Suppose g is continuous at x0 and g(x0) =£ 0. By Lemma 3.3, there are 8 > 0 and a > 0 such that for \x - x0\ < 8 and x E D, we have \g(x)\ > a. Choose €> 0. Let €f =

a\g(xo)\€>0.

There is 8f > 0 such that for |JC - x0\ < 8' and

xGD,

\g(x) - g(x0)\ < €'. Let 8" = min{5', 8}. Now 8" > 0 and for \x - x0\ < 8" with x E D, we have - ( * ) - - (*>)

g(*b) ~ g W a|g(^o)|

Thus, j is continuous at *0* By item (ii) of this theorem, | is then continuous at x0. From our remarks in Chapter 2 and Theorem 3.1, it is clear that every polynomial is continuous. Furthermore, if p and q are polynomials, then^ is continuous at any point x where q{x) # 0. The case where q(x) 0 is handled in detail in Exercise 12 at the end of this chapter. • Example 3.1 The function fix) = sin x is undoubtedly familiar to the reader. We observe that, for 0 < x < w/2, 0 < sin x < x. From this and the fact that

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88 Chapter 3 Continuity sin(-x) = -sin x, it is clear that f(x) = sin x has a limit at zero, namely, lim^o sin x = 0. From this, we can deduce that the sine function is continuous at every point. Now for all x and y, x +y 2 / sin """ \

sin x — sin y = 2 cos

2

Let x0 he any real number and € > 0. Let 8 = min{ €, TT] . Then for \x0 — y | < 6, I sin JC0 - sin y\

2 cos ^ — j

sm

^ — j 0 and let 8 - min{ e, 7r}. If |* - x0\ < 8, then

( ! - ' ) - ( ! - * )

= |*6 - *| < 8;

hence, | cos x — cos x01 = s , n | - - * l - s i n l -

XQ

< 8 < €.

Thus, cos JC is continuous at *0- Now it is clear that the functions tan xt sec x, esc x, and ctn JC are continuous where defined, since they may be defined by appropriate products and quotients of the functions sin x and cos x. • Two points that came to light in the preceding paragraph deserve some special attention. First, a function / : D ~> R is continuous at a point x0 E D iff for each e > 0, there is 8 > 0 such that for \x - x0\ < 8 and x E D, we have |/(x) - /(^ 0 )l < *• Note that the choice of 8 > 0 is influenced by two things: the choice of e > 0 and the point in question, x0. In the process of investigating the continuity of f(x) = sin JC, it was discovered that if €> 0 is given, 8 > 0 could be found that would be suitable for any JC0. This is not the usual state of affairs and will be considered in some detail very soon. Second, the continuity of the cosine function was deduced from the facts that the sine function is continuous and that cos JC = sin(f - x). If we let g(x) = cos x, f(x) = sinx, and h{x) = f - JC, then this identity may be written g = / ° h. Of course, since h is a polynomial function, it is continuous; indeed, that fact was used to observe that if |JC — jc0| < 8, then \h(x) - h{xQ)\ =

IT

~ X

~

1T

*0

=

|JC

- jc0| < 8.

We seek now to generalize this idea and shall attempt to prove that the composition of continuous functions is continuous.

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33

Uniform Continuity: Open, Closed, and Compact Sets 89

3.4 THEOREM If / : D -> R and g : D' -» R with im / C D', where / is continuous at x0 E D and g is continuous at /(x0), then g ° / is continuous at XQ. Proof Choose e > 0. There is 5, > 0 such that if \y - f(x0)\ < 5, and y E D\ then

\g(y) ~ *(/(*>)) I < e. There is S2 > 0 such that if |JC - x0\ < ^ and or E A then | / t o - f(xo)\ < «i. Now, since im / C D', if |JC - x0\ < 82 and x E D, then / t o E D\ f(x0) E D', and | / t o - / ( A T 0 ) | < 5,; hence,

|(g*/)to - (g*fXxo)\ = l*(/to) - £(/(*o))| < e Therefore, g ° / is continuous at x0. This theorem, coupled with previous results, allows us to conclude immediately that functions such as / t o = cos x2 and g(x) = sin(cos JC) are continuous at x0 for all real numbers x0> Let us now take up the remaining topics that we want to pursue further. We shall give the appropriate definition and then consider some examples.

3.3 UNIFORM CONTINUITY: OPEN, CLOSED, AND COMPACT SETS DEFINITION A function f:D-*Ris uniformly continuous on E C D iff for every e > 0, there is 5 > 0 such that if x, y E E with \x - y) < 8, then I/to ~ /(y)| < €. If / is uniformly continuous on D, we say / is uniformly continuous. As promised, some examples and nonexamples are forthcoming. •

Example 3.2 Consider the function / : [2.5, 3] -» R defined by /(*) =

3

.

We will show that / is uniformly continuous on [2.5, 3]. Choose 6 > 0. Now JKy)

x-2

y-2

0t-2)(y-2)'

We see no difficulty making the numerator of this last fraction small, but, as we have seen before, the problem of bounding the denominator away from zero must be addressed. However, since x and y must belong to [2.5,3], the smallest nonnegative value

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90 Chapter 3 Continuity

for the denominator is 0.25, occurring when x = y = 2.5. With this in mind, we choose 8 = -fee. Thus, if \x - y\ < S and x, y G [2.5, 3], then \f(x) - f(y)\ =

3(y - x) (x - 2)(y - 2)

0.2 5

• Example 3.3 Consider the function / : (0,6) -*'J? defined by f(x) = x2 + 2x - 5. We will show that / is uniformly continuous on (0, 6). Choose €> 0. Now \f(x) - f(y)\ = \{x2 + 2x - 5) - (J2 4- 2y - 5)| = \(x2 - y2) + 2(x - y)| = |(^-y)(x + y + 2)|. To make this expression small, we need to make \x - y | small and bound \x + y + 2|. Although 6 is not in the domain of /, we can use that number to bound |JC + y + 2|; that is, |JC + y + 2| < 6 4- 6 + 2 = 14. With this in mind, we choose 8 = -fee. Thus, if JJC — y | < 8 and JC, y E (0, 6), then

1/00 - f(y)\ = Ifr - y)fr + y + 2)| < 14a = e

•

The sine function has been shown to be uniformly continuous on the set of all real numbers. Therefore, it is a uniformly continuous function. Similarly, the cosine function is uniformly continuous. This is not surprising in light of Exercise 23. • Example 3.4 Consider the function / : (0, 1) -> R defined by f(x) = 1/JC. To show that this function is not uniformly continuous, we will choose € = 1 and show that for any 8 > 0, there are x and y in (0, 1) such that |JC — y\ < 8 and 1/00 "" f(y)\ — 1- Suppose 8 > 0. Let a = min{2, 8}, x = a/3, and y = a/6. Then |* - y\ = a/6 < 8, and \f(x) - f(y)\ = | | - | | = | | | ^ 1. • Now consider the function / : (0, IT/2) —» R defined by f(x) = tan JC. Since tan JC = sin JC/COS x, for 0 < x < TT/2, f is a quotient of uniformly continuous functions. (Since sin JC and cos JC are uniformly continuous on R, they are certainly uniformly continuous on (0, TT/2).) Exercises 19 and 20 show under what conditions the sum, product, or composition of uniformly continuous functions yields a uniformly continuous function. We know that the tangent function is unbounded near TT/2, SO it cannot have a limit there. Let us show that this guarantees that tan JC is not uniformly continuous on (0, TT/2).

3.5 THEOREM Let / : D ~> R be uniformly continuous. Then, if JC0 is an accumulation point of D, f has a limit at JC0. Proof Let JC0 be an accumulation point of D, and let {JC„}*=1 be any sequence of members of D\ {x0} converging to JC0. Recall that it is sufficient to prove that {f(xn)}n=t is a Cauchy sequence, since every Cauchy sequence of real numbers is convergent. Choose €> 0. Since / is uniformly continuous on Z), there is 5 > 0 such that for all JC, y E D, \x - y | < 8 implies \f(x) - /(y)| < & Since {*„)"..i converges to JC0, it is a Cauchy sequence, and there is N such that for m, n ^ N,

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3.3 Uniform Continuity: Open, Closed, and Compact Sets 91

|JC„ - xm\ < 8. Now xn E D for each n; hence, for m,n>N, \xn - xm\ < 8 and xm, xn E D; hence | / W - f{Xm)\ < 6. Thus, {f(xn)}n=i is Cauchy. By Theorem 2.1, / has a limit at x0. m Example 3.5 Observe that Theorem 3.5 gives a necessary condition for uniform continuity, but not one that is sufficient. For example, consider g:R -» R where g(x) = x2 for all real numbers x. Since g is continuous and dom g = R, the function g has a limit at every accumulation point of R because R contains all its accumulation points. However, \g(x) - g(y)\ = k 2 ~ y2\ = \x + yll* - j | Choose e > 0 and consider any 8 > 0. Now choose x and y such that |* - y\ = 5/2 < 8 and |JC + j | = 3e/& Then IsC*) - 8(y)\ = \x + j l k - y\ = j 2 = 2 € > 6* If effect, given €> 0, it is impossible to find 5 > 0 such that for |JC - y\ < 8, we always have \g(x) - g(y)\ < e. Thus, g is not uniformly continuous. • We have seen two examples of continuous functions that were not uniformly continuous; in the first case, the domain was (0, f)—a set with accumulation points that did not belong to the set—and the second function had an unbounded domain. The difficulty in the first example comes about near the accumulation points and perhaps may be overcome by considering functions whose domains contain all their accumulation points. Such sets are of sufficient importance to deserve a name. DEFINITION A set E C R is closed iff every accumulation point of E belongs to£. As has been observed previously, afiniteset has no accumulation points and, by definition, is a closed set. Also, Rf the set of all real numbers, is a closed set, since an accumulation point of R is necessarily a real number, hence a member of R. Suppose £ is a closed set and x0 E R\E. Then, by the definition of a closed set, x0 is not an accumulation point of E; hence, there is a neighborhood Q of x0 that contains no points of E (since x0 &E\$oQCR\E. Thus, if £ is closed, there is, for each x0 E R\E, a neighborhood Q of x0 such that Q C R\E. This special type of set comes about in a natural way by considering the complement of a closed set. DEFINITION A set A C R is open iff for each x E A there is a neighborhood Q of x such that Q C A. The remarks leading up to this definition yield the proof of half of the following theorem.

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92 Chapter 3 Continuity 3.6 THEOREM

A set E C R is closed iff R \E is open.

Proof Suppose E is closed and x0 E R\E. Then, since £ contains all its accumulation points, x0 is not an accumulation point of E. Hence, there is a neighborhood Q of x0 that contains no points of E. (Recall that x0 $ E.) But now Q C R\E; hence (by definition), R\E is open. Suppose R\E is open. To show E is closed, it suffices to show that if x0 is an accumulation point of E, then x0 E E. Let x0 be an accumulation point of E. Then x0 E E or x0 E R\E. However, R\E is open; so if JC0 E / A £ , there is a neighborhood Q of x0 such that 2 C / ? \ £ o r < 2 n £ i s empty, contrary to x0 being an accumulation point of £. Thus, x0E E and £ is closed. Now let us return to the problem that precipitated this discussion. Consider f :E-> R with / continuous on £. Choose €> 0. Now for each xE Ef there is 8X > 0 such that if y E E and \x - y\ < 8X, then \f(x) - f(y)\ < e. In the search for 8 > 0 to guarantee that |JC — y\ < 8 and x, y E E imply that \f(x) - f(y)\ < e, one might consider 8 = M{8x:xE

E}.

If E is finite, 8 is positive; but if E is infinite, 8 might be zero. Indeed, if / is not uniformly continuous, it may be zero. Consider now the set (x - 8x,x + 8X) n E. If yu y2E(x

- 8X, x + 8X) D E,

then \x - yx\< 8X and \x — y2\ < 8X; hence, l/(yi) " f(yi)\ =s |/(y0 - / « + /(JC) -

f(y2)\

* \f(yi) - / W l + l / t o - /(y?)! < € + € = 2 €. If one could find jcb . . . , xn such that n

E C |J (*, - «,,, x, + 8X), 1=1

there might be some hope of showing that / is uniformly continuous. Note that for each x, (x — 8X, x + 8X) is an open set (see Exercise 27), and E C UJCE£(JC — 8A., x 4- 8X). Here we have a family of open sets whose union contains E, and we wish to choose a finite subfamily with the same property. A new definition is now in order. DEFINITION A set E is compact iff, for every family {Ga} aeA of open sets such that E C Ua(EAGa, there is a finite set {a„ . . . , an] C A such that E C UJL,Ga, We shall illustrate this concept by giving some examples of sets that are not compact.

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3.3 Uniform Continuity: Open, Closed, and Compact Sets 93 • Example 3.6 Let E = (0, 1] and for each positive integer n, let Gn = (£, 2). If 0 < x ^ 1, there is a positive integer n such that ~ < x; hence, x E Gn, and thus 00

EC{J

G n.

If we choose a finite set « i , . . . , nr of positive integers, then

U 0„,. = G„0

1=1

where n0 = maxf/^,..., /ir} and

Thus, we have a family of open sets {Gn}n&J such that E C U we/ G w , but no finite subfamily has this property. From the definition, it is clear that E is not compact.

• Example 3 J As a second example, consider the set / of all positive integers. For each « G / ( define

G n H

'-{ -h

+

i)-

Now / C U~=,GW, but each Gn contains exactly one member of J, so it is impossible to choose n i , . . . , nr such that / C Wi=lGnr Thus, / is not compact. • Let us introduce some new terminology to help simplify our language. If £ is a set, and {Ga}aeA is a collection of sets such that E C UaeAGa, then the collection {Ga)aeA is called a cover of £. If each Ga is an open set, then the collection {Ga}aeA is called an open cover of E. If {Ga)a&A is a cover of E and B C A such that E C UaBBGa, then the collection {C t t } a e B is called a subcover of E; and if £ is finite, then {Ga}aeB is called &finitesubcover of £. In a sense, we are abusing the language, since every subcover of £ is also a cover of £; but in light of the notion of compactness, this is a reasonable abuse. With these additions to our vocabulary, we can restate our definition of compactness: A set £ is compact iff every open cover of £ has a finite subcover. The reader should keep the preceding two examples clearly in mind. The first example, (0, 1], is a bounded set that is not closed, since 0 is an accumulation point of (0,1 ] that does not belong to (0,1]. The second example is a closed set but is unbounded. As the reader might suspect from our choice of examples, the compact sets on the line are precisely those that are both closed and bounded. This is the content of the next theorem.

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94 Chapter 3 Continuity 3 J HEINE-BOREL THEOREM bounded.

A set E C R is compact iff E is closed and

Proof First we prove that any closed interval is compact. Let [a, j8] be a closed interval and {GA}AeA be an open cover of [a, /?]. There is A0 E A such that a E GAo. Hence there is e > 0 such that (a - e, a 4- e) C GAo. This means that the interval [a, a + f ] can be covered by the open set G^. Let £ = {x: JC E [a, 0] and [a, JC] can be covered by a finite subfamily of {GA}AeA}. B is bounded and nonempty since [a, a + § ] C B. Let z = sup B. We know that a + f ^ z and z < /3. So, there is \x E A such that z E GXl. Hence there is 5 > 0 such that (z - 8, z + 8) C GAl. Now [a, z - f] can be covered by a finite subfamily of {GA}AeA and hence [a, z + f] can be covered by the subfamily of {G A } A G A that covers [a, z - f ] plus the open set GAj. This contradicts the definition of z unless z = /3. Thus B = [a, j8] and [a, /3] is compact. Now suppose E is closed and bounded. Then there is a closed interval [a, j3] such that E C [a, /3]. Let {G a } a6i4 be an open cover of E. Then we obtain an open cover of [a, j3] by adding the open set R\E to the family {Ga}a€=A of open sets that cover E. Since [a, /3] is compact, there is a finite set {au a2,..., a„} such that G tt| , G « 2 , . . . , Gan together with R\E is an open cover of [a, 0]. But E C\ R\E = 0. Since / is continuous on D, f is continuous at x for each x G D. Thus, for each x E D, there is 8X > 0 such that if | JC - y \ < Sx and y E D, then |/(x) - f(y)\ < f. Consider this family

2/LD This is an open cover of D and D is compact, so there is a finite subcover of D. In other words, there are xu . . . , xn E D such that

DCU

X

' - l

, Xi 4-

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96 Chapter 3 Continuity

Let o = mm i — : i = 1 , . . . , n K Now suppose x, y E D and \x - y| < 8. There is an integer / among 1 , . . . , n such that

8Xi Now |JC — y| < 5 s —I, making

\y - */l ^ |y - *l + k - *i\ < y + y

=

^

Hence, |/(x) - f(y)\ ^ |/(x) - /(*,)| + l/fe) - / 0 » | < | + | - e. Therefore, / is uniformly continuous. In a broad sense, this theorem has a converse. Suppose / : D -» R is uniformly continuous. As has been pointed out, / has a limit at each accumulation point of D, As in Exercise 28, let D' be the set of accumulation points of D andD = D U D ' . Define g : D -* R by #(*) = /(r) for all x G £>; and if Xo G D \ D , define g(;c0) = lim*^ /(*). The function g : £> ~* # is called an extension of / : £> -»J? and indeed turns out to be a continuous function. (The proof of this will be a challenging exercise for the reader.) Thus, if / : Dj-» R is uniformly continuous with D bounded, then / has a continuous extension g :D -> R whereD is closed and bounded and hence, compact.

3.4 PROPERTIES OF CONTINUOUS FUNCTIONS Some of the reasons for considering continuous functions are the special properties they possess. In the next few pages we consider some of these properties and their applications. 3.9 THEOREM If / : E -» R is uniformly continuous and £ is a bounded set, then f(E) is a bounded set. Proof Choose 6 = 1 . There is 8 > 0 such that, if |JC - y| < 8, then \f(x) - f(y)\ < 1. Since E is a bounded set, there are xu x2,...»xn in E such that EC\)

& - 8, xt + 8),

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3.4 Properties of Continuous Functions 97 but then n

f(E) C [J (/(*.) - I- /(*.) + D. 1=1

since |JC - xt\ < 5, x E £, implies that \f(x) - / f o ) | < 1. Thus /(£) is bounded. Suppose f ;E->R is continuous with E compact. Then / is uniformly continuous and E is bounded, so f(E) is bounded. If we could prove that f(E) was closed, then f(E) would be compact, and we could say that the continuous image of a compact set is compact. Before proceeding, we need the following lemma. LEMMA Let E be a compact and {xn}~=1 be a sequence of points in E. Then there is x0 in E and a subsequence {x„k)Z=i that converges to x0. Proof Since E is compact, it is both closed and bounded. Thus {*n}"=1, being bounded, has a convergent subsequence {xnk\n=\. The limit of this subsequence JC0 belongs to £ or is an accumulation point of £. (See Theorem 1.17.) But £ is closed, hence JC0 E £. 3.10 THEOREM compact.

Let / : £ - » / ? be continuous with £ compact. Then /(£) is

Proof As pointed out in the preceding paragraph, we need only show that /(£) is closed, since we already know /(£) is bounded. Let y0 be an accumulation point of /(£). Then there is a sequence {y„}"«i of points in / ( £ ) such that yn ¥=• v0 for all n and {yn)n=i converges to y0 (see Theorem 1.17.) Since yn E / ( £ ) for each n, there is a sequence {JC*}^! of points in £ such that f(xn) = yn for each n. There is a subsequence U„jr=i that converges—call the limit x0—and x0 E £ by the lemma. The sequence {f(xnk)}1°=l converges to f(x0), since / is continuous at x0. But

is a subsequence of {jw}n=i and converges to y0, so j 0 = f(x0) E /(£). Therefore, /(£) is closed. The following important corollary follows from Theorem 3.10. 3.11 COROLLARY If / : £ ~* R is continuous with £ compact, then there are xu x2 in £ such that, for all x E £,

Proo/ By Theorem 3.10, /(£) is a compact set, and then, by Exercise 35, both sup /(£) and inf /(£) belong to /(£). Thus, there are jq and x2 in £ such that f(xx) = inf / ( £ )

and

f(x2) = sup /(£).

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98 Chapter 3 Continuity

Therefore, for all x in E, f(xx) * f(x) R but is 1-1 but such that f~l : f(E) -» E is not continuous. Of course, E must be noncompact. Let E = [0, 1] U [2, 3), and define f(x) = JC for 0 < x x

Figure 3.3

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100 Chapter 3 Continuity

Now consider the sequence

which converges to 1. For n odd, ii,.

( - D " \ _ , . (-D"

and for n even,

r_/

It should be clear then that if fails to be continuous at 1. 3.5.

l

I1H

(-ir\r I>

does not converge, and so /

!

•

If you wish to discover more about the continuity of inverse functions, see Project

Consider now a function / : [-1, 1] -» R that is continuous and is such that / ( - 1 ) < 0 and /(I) > 0. An intuitive picture of the geometric nature of the graph of a continuous function leads us to guess that this graph must cross the x axis somewhere between - 1 and +1; that is, there is x E ( - 1 , 1) such that f(x) = 0. This indeed turns out to be the case; in fact, it motivates us to state the following theorem, published by Bernard Bolzano in 1817. 3J3 BOLZANO'S THEOREM Suppose / : [a, b] -» R is continuous and f(a) and f(b) have opposite signs. Then there is z E (a, b) such that /(z) = 0. Proof Consider the case where f(a) < 0 and f(b) > 0, and define three sequences {*X-i. {?X-i. and {cX=i as follows: X\ = a, y, = bt c, = \{xx + J,), and for n ^ 1, 1. if f(cn) < 0, define x„+l = cn and yn+l = y„. 2. if f(cn) > 0, define xn+x = xn and yn+l = cn. Notice that for each n, cnt xnf and y„ are in the interval [a, b], hence they belong to the domain off. Now, {xn}~=l is an increasing sequence, { y X ^ is a decreasing sequence, both are bounded and hence they both converge. Moreover, \xn+l - yn+l | < 2~"(b - a) for all n. Therefore both sequences converge to the same limit; call it z. Both {f(xn)} "=, and {f(yn)} "=, converge to f(z) since z belongs to the domain of / and / is continuous at z. But f(yn) ^ 0 for each n and f(xn) < 0 for each n, hence /(z) = 0. The case for f(a) > 0 and f(b) < 0 follows easily.

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3.4 Properties of Continuous Functions 101

As a beginning student of analysis, you should cultivate the practice of dissecting and examining a proof to determine those facts essential to that proof. Such examination of this proof yields a big payoff in terms of a much more general theorem with almost no effort. Notice that it wasn't important that the domain of / be an interval, but rather that any points between a and b belong to the domain of / . We formalize this idea in the following definition. DEFINITION A set A C R is connected if whenever a and b are in A and a < c < b, then c E A. The connected sets in R are fairly simple to characterize. The following theorem accomplishes that task. 3.14 THEOREM following:

Let A be a connected subset of R. Then A is one of the

i. {x :x < a}, (x: x > a), (x: x ^ a}, {x : x ^ a] ii. [a, b], [a, b\ (a, b], (a, b) iii. R Proof It is easy to check that each of the sets listed is connected. So we need to show that if A is connected, then it falls into one of the three categories. Note that the empty set is connected and is included in category (ii). Do you see how? Now assume A is nonempty. If A is neither bounded from above nor from below and x E R} then x is neither an upper bound nor a lower bound for A. Hence there are a, b E A such that a < x < b. Since A is connected, x E A. Hence A = R. Assume A is bounded from above, but not from below, and let a = sup A. If x E R, since x is not a lower bound for A there is v E A such that y < x. If x < a, then there is z E A such that x < z < a. Thus y < x < z and since A is connected, x E A, Thus A — {s : s < a] or A = {s : s ^ a}. We leave the remaining cases to the reader. If you have more interest in connected sets, look to Project 3.4. Now we can state a modified version of Bolzano's Theorem. 3.15 MODIFIED BOLZANO'S THEOREM Suppose / : A -* R is continuous with A connected. Then if a < b, a, b E A with f(a) and f(b) having opposite signs, there is c E (a, b) such that /(c) = 0. Proof

The proof of Theorem 3.13 may be used without changes.

Now we are prepared to state and prove the Intermediate-Value Theorem.

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102 Chapter 3 Continuity

3.16 INTERMEDIATE-VALUE THEOREM Let f:A -» R be continuous with A connected. Further suppose a < b, a, b E A with f(a) < y < f(b) (or f(a) >y> f(b)). Then there is c E (a, 6) such that f(c) = y. Proof Define g : A -»/? by g(jt) = f(x) - y. Then g is continuous and g(a) and g(b) have opposite signs. Consequently, by the modified Bolzano's Theorem, there is c E (a, b) such that g(c) = 0. Therefore 0 = g(c) = f(c) - y, and /(e) = y. Theorem 3.10 states that if / \D -» # is continuous and D is compact, then /(£>) is compact. In other words, compactness is a property of a set that is preserved by a continuous function. Note that the Intermediate-Value Theorem states that if / : A -» R is continuous and A is connected, then f(A) is connected. We conclude that connectedness is also a property of a set that is preserved by a continuous function. The Intermediate-Value Theorem is, of course, the one that is used in trying to find the zeros of a polynomial. If p is a polynomial with p(a) < 0 and p(b) > 0, then p has at least one zero between a and b, since every polynomial is continuous. If p is a polynomial of odd degree, then p will have a change of sign, so it necessarily has a zero. Let us prove this fact. 3.17 THEOREM If p is a polynomial of odd degree with real coefficients, then the equation p(x) = 0 has at least one real root, Proof Assume p(x) = a0 + axx + • • • + a^cn9 where n is odd and an -=h 0, If a0 = 0, then p(0) = a0 = 0 and we are through. Assume now that a0 # 0, and let

for all x # 0. Choose € such that 0 < e < \an\. For i - 0, 1, 2 , . . . , n - 1, there exists &/ > 1 such that

i£d 0 with JC > £, then p(x) > 0; and, if x < -JC, then p(x) < 0. Similarly, if

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3.4 Properties of Continuous Functions 103 an < 0 and x > K, then p(x) < 0; and, if x < -if, then p(x) > 0. In any case, we can find points xt and x2 such that p{xx) < 0 and p(x2) > 0; hence, somewhere between xx and JC2, p has a zero. If we combine Corollary 3.11 and Theorem 3.16, we obtain the following result. 3.18 THEOREM

If / : [a, ft] -> R is continuous, there are c and rf such that

ma,b]) = [c,d\. Proof By Corollary 3.11, there are xx and x2 in [a, 6] such that f(xx) < /(or) < f(x2) for all JC E [a, ft]. Let c = f(xx) and rf = /(x 2 ). By the Intermediate-Value Theorem, if c ^ y ^ d, there is x E [a, b] such that /(JC) = y. Thus /([a, ft]) = [e, dj. • Example 3.11 As a further application, we shall show that the equation x = cos x has at least one solution in the closed interval [0, f ]. Consider the function f(x) = x - cos x on [0, f ]. Clearly, / is continuous, /(0) = - 1 , and / ( f ) = f. Therefore, there is x E [0, f ] such that /(JC) = 0; in other words, x = cos x. • 3 J 0 THEOREM / is monotone.

Let / : /4 -> /? be continuous and 1-1 with i4 connected. Then

Proof Suppose / is 1-1 and continuous but not monotone. Then one of the following cases applies: i. there are x, y, z E A such that x < y < z and /(*) < f(y) and /(z) < /(y) or ii. there are x, y, z E A such that x < y < z and f(x) > f(y) and /(z) > f(y). Let us consider case (ii). We have here x, y, z E A, x < y < z and f(x) > f(y) and /(z) > /(y). Let us now suppose that /(y) < /(z) < /(JC). By the IntermediateValue Theorem, there is w E [JC, y] such that f(w) = /(z) contrary to / being 1-1. The proof for the other cases is left to the reader. Let us now return to a topic considerecj^at the end of Chapter 2. Let / : [a, j8] ~» R be monotone. Then D = {x: JC E (a, j8) and / does not have a limit at JC} is countable; and if / has a limit at x0 E (a, 0), then lim^^C*) = /(JC 0 ). The discussion of the behavior of / at a and at 0 was left to the reader. It is no betrayal of confidence to state now that / has a limit at a and at /3 and that lim /(JC) = inf{/(jc): a < JC < j8}

x-*a

and lim f(x) = supf/Oc): a < JC < /3]

x-0

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104 Chapter 3 Continuity if / is increasing. In the case where / is decreasing, both limits still exist, but lim /(*) = sup{/(x): a < x < 0} and lim f(x) = inf{/(jc): a < x < 0}. Now it is clear that / is continuous at x for each x E (a, f3)\D, and possibly at a and £ depending on the definition of / at a and /3. In particular, the set of points at which / is discontinuous is countable.

EXERCISES 3.1 CONTINUITY OF A FUNCTION AT A POINT 1. Define / : / ? - » R by f(x) = 3x2 - 2x + 1. Show that / is continuous at 2. 2. Define / : [-4, 0] -* /? by /(JC) =

9y2 _ JC

3. 4. 5. 6. 7. 8. 9. *10. 11.

JO

— for x ± - 3 and /(-3) = -12. Show that / is

+ 3

continuous at - 3 . Use Theorem 3.1 to prove that f^V!+1)/?=i is convergent, and find the limit. You may assume that the function f(x) = ex is continuous on R. If xQ E E, X0 is not an accumulation point of E, and f :E-»R, prove that, for every sequence {**}£= i converging to JC0 with JC„ G E for all «, {/(*„)}*=, converges to f(x0). Define / : (0, 1) -» /? by /(*) = -7= - / — — . Can one define /(0) to make / continuous VJC \ x at 0? Explain. Prove that /(*) = Vx is continuous for all x ^ 0. Suppose / : R -> R is continuous and /(r) = r2 for each rational number r. Determine /(V2) and justify your conclusion. Suppose f :(a,b)->R is continuous and f(r) — 0 for each rational number r G (a, &). Prove that f(x) = 0 for all x G (a, b). Define / : (0, 1) -» /? by /(JC) = x sin 7. Can one define f(0) to make / continuous at 0? Explain. Suppose f :E-* R is continuous at x0, and x0 E F C E. Define g : F -» /? by g(x) = /(JC) for all xEF, Prove that g is continuous at x0. Show by example that the continuity of g at JC0 need not imply the continuity of / at JC0. Define f:R->Rby f(x) = 8* if x is rational and f(x) = 2x2 4- 8 if JC is irrational. Prove from the definition of continuity that / is continuous at 2 and discontinuous at 1.

3.2 ALGEBRA OF CONTINUOUS FUNCTIONS *12. Let p and q be polynomials and x0 be a zero of q of multiplicity m. Prove that p/q can be assigned a value at JC0 such that the function thus defined will be continuous there iff x0 is a zero of p of multiplicity greater than or equal to m. 13. Let / : D -» R be continuous at x0 E D. Prove that there is M > 0 and a neighborhood Q of JC0 such that |/(x)| < M for all JC G Q fl D. 14. If / : £> -> /? is continuous at JC0 G D, prove that the function \f\:D -> # such that I/lOO = t/Wi is continuous atx0.

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Exercises

105

15. Suppose / , g : D -> R are both continuous on ZX Define h : D -> R by /*(*) = max {/(*), g(x) (. Show that h is continuous on D. 16. Assume the continuity of f(x) = e* and g(jc) = In JC. Define /*Cc) = xx by JC* = xxXnx. Show & is continuous for x > 0. 17. Suppose f :D -* R with /(JC) > 0 for all J C 6 D . Show that, if / is continuous at JC0» then V J is continuous at JC0. 18. Define f:R-+Ra$ follows: /(*) = A* - [JC] if [x] is even. /(*) = x - [x + 1] if M is odd. Determine those points where / is continuous. Justify. 3.3 UNIFORM CONTINUITY: OPEN, CLOSED, AND COMPACT SETS 19. Let / , g : D —> R be uniformly continuous. Prove that / -4- g : D —» R is uniformly continuous. What can be said about fgl Justify. 20. Let / :A —> £ and g : £ —> C be uniformly continuous. What can be said about g ° / M - » C ? Justify. 2 21. Define / : [3.4, 5] -» R by /(*) = _ , . Show that / is uniformly continuous on [3.4, 5] without using Theorem 3.8—that is, use the methods of Example 3.2. 22. Define / : (2, 7) - » R by f(x) = JC3 - x + 1. Show that / is uniformly continuous on (2, 7) without using Theorem 3.8—that is, use the methods of Example 3.3. 23. A function / : R —» R is periodic iff there is a real number h # 0 such that /(JC + h) = f(x) for all x G /?. Prove that if f :R -> R is periodic and continuous, then / is uniformly continuous. 24. Suppose A is bounded and not compact. Prove that there is a function that is continuous on A, but not uniformly continuous. Give an example of a set that is not compact, but every function continuous on that set is uniformly continuous. 25. Give an example of sets A and B and a continuous function f :A U B -* R such that / is uniformly continuous on A and uniformly continuous on B, but not uniformly continuous on A\JB. *26. Let E C R. Prove that E is closed if, for every x0 such that there is a sequence {*„}"=! of points of E converging to x0, it is true that x0 G E. In other words, prove E is closed if it contains all limits of sequences of members of E. *27. Prove that every set of the form [x: a < x < b] is open and every set of the form {JC : a ^ JC ^ b) is closed. 28. Let D C R, and let D' be the set of accumulation points of£>. Prove that D = D U D' is closed and that if F is any closed set that contains D, then D C F. D is called the closure ofD. _ 29. If D C R is bounded, prove that D is bounded. 30. Suppose f :R-*R is continuous and let r0 E R. Prove that (x : /(JC) =£ r0} is an open set. 31. Suppose / : [a, b] —> /? and g : [a, fr] -> /? are both continuous. Let T = {JC : /(x) = g(jc)}. Prove that T is closed. 32. IfDCR, then JC G D is said to be the interior point of D iff there is a neighborhood Q of x such that Q C D. Define £>° to be the set of interior points of D. Prove that D° is open and that if S is any open set contained in D, then 5 C D°. D° is called the interior of D. 33. Find an open cover of {JC : x > 0} with no finite subcover. 34. Find an open cover of (1, 2) with no finite subcover.

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106 Chapter 3 Continuity *35. Let E be compact and nonempty. Prove that E is bounded and that sup E and inf E both belong to E. 36. If / ? ! , . . . , En are compact, prove that E = U/LiZs, is compact. 37. Let / : [a, b]-> R have a limit at each x G [a, &]. Prove that / is bounded. 38. Suppose / : D —> R is continuous with Z) compact. Prove that {JC: 0 ^ /(JC) ^ 1} is compact. 39. Suppose that f :R -* R is continuous and has the property that for each € > 0, there is M > 0 such that if | JC | ^ M, then |/(JC)| < e. Show that / is uniformly continuous. 40. Give an example of a function f :R -* R that is continuous and bounded but not uniformly continuous.

3.4 PROPERTIES OF CONTINUOUS FUNCTIONS 41. Find an interval of length 1 that contains a root of the equation xex = 1. 42. Find an interval of length 1 that contains a root of the equation x3 - 6x2 + 2.826 = 0. 43. Suppose / : [a, b] -> R is continuous and fib) < y < f(a). Prove that there is c G [a, b] such that /(c) = y. 44. Suppose that / : [a, b] -» [a, b] is continuous. Prove that there is at least one fixed point in [a, b)—that is, x such that /(JC) = x. 45. If/ : [a, 6] -» R is 1-1 and has the intermediate-value property—that is, ify is between /(«) and /(i>), there is JC between u and u such that f(x) = y—show that / is continuous. (Hint: First show that / is monotone.) 46. Prove that there is no continuous function f :R-*R such that, for each c G R, the equation /(JC) = c has exactly two solutions.

MISCELLANEOUS 47. Let f:R -* R be additive. (See Project 2.1 at the end of Chapter 2.) That is, f(x + y) = /(JC) + fiy) for all JC, y G /?. In addition, assume there are M > 0 and a > 0 such that if x G [- /? be continuous, and define g : [A, b] -> /? by g(0 = sup{/(x): R with x0 an accumulation point of D and x0 G D. For each x G D with x =£ x09 define

m=

fix) ~ /(*o) X —

XQ

The function / is said to be differentiable at x0 (or has a derivative at x0) iff T has a limit at JC0, and we write lim*.^ T(x) = f(x0). The number f(x0) is called the derivative of / at JC0. If / is differentiable for each x G E C D, we say / is differentiable on E. The following alternative definition of the derivative may also be familiar. In Exercise 2, you are asked to prove that the two definitions are equivalent. ALTERNATE DEFINITION Let / : D -* R with x0 an accumulation point of D and JC0 G £>. For each t G R such that ;c0 + t G D and t # 0, define =

/ f a + 0 - /(XQ)

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4.1 The Derivative of a Function 113 The function / is said to be differentiable at xQ iff Q has a limit at zero. If this limit exists, it is called the derivative off at x0. Now we shall state a basic theorem that gives a condition for differentiability in terms of sequences. The definition of differentiability is based on the existence of the limit of a function, so this theorem should come as no surprise. 4.1 THEOREM Suppose / :D -» R,x0E D, and x0 is an accumulation point of D. Then / is differentiable at x0 iff for every sequence {xn} ~= x of points of D \ {x0} converging to x0, the sequence

[

xn - x0

Jw==1

converges. No formal proof will be supplied here because it should be clear from Theorem 2.1 that the condition concerning sequences is equivalent to the condition that the function T given in the definition of differentiability has a limit at x0. Note also that if / is differentiable at x0 and {xn}n=i is a sequence of points of D\{x0] converging to x0, then I

xn - x0

)nwml

converges to f(x0). The next few examples will further one's understanding of the definition of differentiation. • Example 4.1 Consider the function f(x) = \x\ for all x E R. Set JC0 = 0, and consider the sequence.

fflj

this sequence converges to zero, but zero is not a term of this sequence. For n even, (-1)"

1

and

n

= 1;

1

n for n odd,

(-D n

1 n

and

H).1 n

/(0) = -1,

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114 Chapter 4 Differentiation

so the sequence /(0)

(-1)" does not converge. Thus, by Theorem 4.1, / is not differentiable at zero. The reader should sketch the graph of |*| to see why this happens. The graph has a' 'sharp" corner at zero with the slope suddenly changing from - 1 to +1 as x increases from negative to positive values, and of course the slope is not defined at zero. • • Example 42 For the moment, assume the usual facts about differentiating the product, sum, and composition of the elementary functions. Define f(x) = x sin £ for x + 0 and f(0) = 0. The fact that / is differentiable at any point other than zero follows from the assumptions made at the beginning of this paragraph. For x =£ 0, /(*) ~ /(0) . -— = s i n x— 0 does not have a limit continuous there.

1 x at zero; hence, / is not differentiable at zero, although it is •

• Example 43 In an attempt tofinda "nicer" function, define f(x) = x2 sin £ for x =£ 0 and /(0) = 0. Again, assume the basic facts about differentiation of products and composition of elementary functions. Now, / has a derivative at x if x =£ 0, and f'(x) = 2x sin j — cos £. So f'(x) is defined for x # 0, but / ' does not have a limit at zero. One might suspect that this means / is not differentiable at 0; however, this is false. For x i1 0, /(*) - f (0) . 1 — = x sin - , x- 0 x which has a limit at zero, namely, /'(0) = 0. In this case we have a function with a derivative at every point, but the derivative is not continuous at zero; in fact, it does not have a limit there. • • Example 4.4 Finally consider f(x) = x3 sin j for x ¥= 0 and /(0) = 0. Again, using the usual rules (to be proven in the next section), / has a derivative everywhere and / ' is continuous everywhere but fails to be differentiable at one point. Can you guess the point? See Exercise 5. • Let us now return to the initial definition and seek the facts pertinent to functions differentiable on a set. First, consider the difference quotient fix) - f(x0) x - x0

=

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4.2 The Algebra of Derivatives 115 For x close to x0, the denominator is close to zero; hence, in order that T have a limit at JC0, the numerator must be close to zero—otherwise T would not be bounded near x0. We are led to predict that, if / is differentiable at x0, then / must be continuous at x0. The proof may be accomplished by observing that *, \ *t ^ f (*) - /(*o) , N f(x) - f(x0) = (x - x0) x - x0 for x T* x0. 42 THEOREM Let / : D -> R be differentiable at jt0. (It is tacitly assumed that x0 E D and that xQ is an accumulation point of D.) Then / is continuous at x0. Let T: D\ {x0] -> R be defined by

Proof

fix) - f(x0)

Tr,

x - x0 T has a limit at x0 and l i n v ^ T(x) = f'(x0). For x i= JC0, f(x) =

(* - ^o) + f(Xo) x - xQ = T(x)(^ - x0) + /(^ 0 ).

Now / is a sum of two functions, one of which is constant and the other the product of two functions, each of which has a limit at x0; so / has a limit at x0 arid lim f(x) = lim [T(x)(x - x0) + f(xo)]

x~*x0

x-*x0

— lim T(x) x-^x0

lim (x - x0)

x-+x0

+ lim f(x0) x-*x0

= f'(x0) • 0 + f(Xo) = f(x0). Thus, / is contirIUOUS at

xQ.

4.2 THE ALGEBRA OF DERIVATIVES The next theorem presents the usual facts from calculus about the sum, product, and quotient of differentiable functions. 43 THEOREM

Suppose / , g : D -> R are differentiable at x0. Then

i. / + g is differentiable at JC0 and (/ + gY(x0) = ff(xo) +

g'M.

ii. fg is differentiable at x0 and (fg)'(x0)

= f(x0)g\x0)

+ /'(Xd)*(Aa).

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116

Chapter 4 Differentiation

iii. If g(x0) # 0, then fig (the domain is the set of all x such that g(x) # 0) is differentiable at x0 and , __ nxo)8(xo) - g'(*6)/fo) [g(X0)f

(fX, W

Proof We shall appeal to Theorem 4.1 to prove this theorem. Let {xn}^\ be any sequence of points in D\ {x0} converging to x0. Since / and g are differentiable atx0, xn - x0 ]n=l

I

converges to f(x0) and

{«\g(xx) -—gC*o)1°° JC J n

w

0

rt=1

converges to g'(jc0). Thus, [(/ + g)fa)-(/^g)(^o)l0° [

Xn — Xo

=

f/fa) - f(x0) + gfa)-gfa>r

Jrt=l

I

Xn — #o

•*« "" *0

J«=i

converges to /'fa) + g'fa)- This means that / + g is differentiable at x0 and that

(/ + *)'fa) = /'fa) + *'(*>). By Theorem 4.2, / is continuous at Jt0, and so {/fa)}~=1 converges to /fa). Thus, hfg)(Xn)

I

_ f/fa)gfa)

- (fg)(X0)y

*» ~ *o

J n =i

[ /(**)

- f(Xn)g(x0)

+ f(Xn)g(Xo)

~ /fa)gfa)1°°

xn - x0

^w -Xb

-r £l*oi —

J w= _

converges to /fa)g'fa) + gfa)/'fa). Thus, fg is differentiable at JC0 and ( M f a ) = f(xo)gf(x0) + gfa)/'fa). Let D' = {x: x E Z) and g(x) # 0}. Then D' is the domain of f/gt and xQ is an accumulation point of D' by the continuity of g at JC0. Let fa}£=i be any sequence of points of D' \ { JC0 } converging to x0. Then, since g is continuous at x0 and g(x0) ¥= 0, 1/g is continuous at x0; hence, {j(x„)}"=i converges to jfa). Now

f//\ /7\ - f a ) - - fe) W

Kg/

Xn =

XQ

f/fa)gfa)

~

gfaXffa)!"

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4.2 The Algebra of Derivatives

117

8(Xn)g(Xo) X

- f(Xo)g(Xo) + f(Xo)g(xQ)

fMgjXp)

~

g(Xn)f(X0)

xn — x0 1

g(*o)

£(*n)«(*b)

f(x„) " /(*b) Xn

XQ

f(x0)

g(Xn) - gfo)) Xn ~~

}'

Jn=l

XQ

converges to 1 r / ^ fe(*b)/'(*o) - f(xo)g,(x0)l Thus, //g is differentiable at x0 and (f\,

s

_ /'(*o)g(*o) - f(x0)g'(x0)

Earlier in this chapter we considered f(x) = x sin \ for JC =£ 0. At that time we assumed some facts concerning the differentiation of the sum, product, and composition of differentiable functions. Now we shall justify our assumptions and consider the differentiability of the composition of differentiable functions. The next theorem is often referred to as the Chain Rule. 4.4 THEOREM (Chain Rule) Suppose / : D -> R and g: D' ~» R with f(D) C D'.lf f is differentiable at *0 and g is differentiable at /(x0), then g ° / is differentiable at x0 and

(r/)'W = ^'(/W)f(4 (Remember that the hypothesis that / is differentiable at x0 implies that x0 is an accumulation point of D, and similarly, g differentiable at /(JC0) implies that f(x0) is an accumulation point of D'.) Proof One might be tempted to prove this theorem by writing (g°f)jx) - (g°f)(xo) X - X0

=

gjfjx)) - g(f(x0)) f{x) - f(x0) f{x) - fix0) f

X - XQ f

and claiming that the right-hand side has g if(x0))f ix0) as a limit, and hence, g ° f is differentiable at x0 as claimed. However, x =£ x0 need not imply that fix) ^ /(*o)- Hence, the denominator may well be zero in the first factor of the right-hand side of the equation. We shall seek to circumvent this trap. Let ^o = f(xo) and, for each y E D', define h(y) =

y - yo

if y # jo

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118 Chapter 4 Differentiation and h(y0) = s'(yo). Since g is differentiable at y0» A has a limit at y0 and lim h(y) = g'(y0) = h(y0). y~*yo

Thus A is continuous aty 0 . Since / is differentiable at JC0andy0 = f(x0), then / i ° / is continuous at x0. The function T defined by T(x) =

for x E D, x =£ JC0 x - x0

has a limit at JC0, namely, /'(JC 0 ). NOW for all x G D, x =£ x0, g(/(*)) ~ g(/(*o)) X XQ

-w N - , v

/L

You should check the case where f(x) = /(JC0). Since both /* © / and T have limits a t % g ° / is differentiable atx 0 and ( r / ) W = lim X-*XQ

X

= hm (h°f){x) XQ

• T(x)

x-*x$

= (*•/)(*>) • f'Qco) = g'(fW)

• /'(*>).

Let us consolidate a few results at this stage. First, any constant function is differentiable and has derivative zero everywhere. The function f(x) = x is differentiable everywhere and fix) = 1 for all x. An easy application of Theorem 4.3 shows that every polynomial function is differentiable and that every rational function (the quotient of two polynomial functions) is differentiable at each point where the denominator is nonzero. Because it is so easy to do and we already know the answer, let us prove now that if f(x) = xn for n an integer, then f(x) - nxn~K (Of course, if n < 0, we must assume x =£ 0, and if n = 0, then fix) - 0 for all x.) 4.5 THEOREM If n is an integer and f(x) = xn for all x, then / is differentiable for all x if n > 0 and for all x * 0 if n < 0, and f(x) = nxn~\ If n = 0, then f{x) = 0 for all x. Proof For n = 0, /(JC) = 1 for all x\ hence, / is differentiable for all x and f(x) = 0, and the theorem is true. We shall now prove the theorem by induction for n > 0 and appeal to Theorem 4.3 to extend to the negative integers. The fact that /(JC) = JCW is differentiable for each positive integer n follows from Theorem 4.3 by induction, but this is the type of induction proof that should be unnecessary for one who has progressed this far. We shall now proceed to show that f(x) = njc*"1. For n = 1, f(x) = JC and fix) = 1 = 1JC°, SO the theorem holds. Suppose the theorem holds for n = r; that is, if fix) - xr, then fix)

= rjcr~l.

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4.3 Rolle's Theorem and the Mean-Value Theorem 119

Consider g(x) = xr+l. We may consider this the product of two functions: f(x) = xr and h(x) = x. By Theorem 4.3, g'M = f(x)hf(x) + f(x)h(x) = x r (l) + (rr r - ! )x = xr + rxr = (/• + l)xr. Thus, the theorem holds for n = r + 1, and the induction is complete. It now remains to verify the conclusion for f(x) = xn for n a negative integer, x # 0. Now, if /? < 0, then -w > 0, and xn = 1/JC"W is differentiable for x # 0 by our remarks earlier in this proof and in Theorem 4.3. Moreover, if f(x) = x" = 1/jTMhen

This concludes the proof. • Example 4.5 Let us examine the use of Theorems 4.3,4.4, and 4.5 infindingthe derivative of a function such as F(x) = V l + 3x2, which is the composition of the function f(x) = 1 + 3x2 and the function g{u) = Vw. For each x, / is differentiable at x and, using Theorems 4.3 and 4.5 f'(x) = 6x. Moreover, for each x, f(x) > 0; hence g is differentiable at f(x). Note that g'(u) = l/2\/u. Thus, by the Chain Rule, g o f = f is differentiable at x and

4 J ROLLE'S THEOREM AND THE MEAN-VALUE THEOREM The classical maxima and minima problems from the calculus are no doubt familiar to the reader. The fact to be applied here is roughly that the graph of a differentiable function has a horizontal tangent at a maximum or minimum point. This is not quite accurate, but let us define some terms before proceeding further. DEFINITION Let / : D ~> R. A point xQ E D is a relative maximum (minimum) of / iff there is a neighborhood Q of x0 such that if x E Q D D, then f(x)^f(x0)

(/to ss/(*>)).

Consider a few examples and reconsider the statement preceding this definition. • Example 4.6 Perhaps the point is most easily made by looking at the function / : [0, 1] —» R such that f(x) = x. Since the domain of / is the interval [0, 1], / has a minimum at 0 and a maximum at 1; indeed, 0 is a relative minimum of /, and 1 is a relative maximum of /. Since f'(x) = 1 for all x E [0, 1], it is clear what is wrong with our statement concerning the existence of a horizontal tangent at a maximum or minimum point. We shall correct this defect in the following theorem. •

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120 Chapter 4 Differentiation

4.6 THEOREM Suppose / : [a, b] —> R and suppose / has either a relative maximum or a relative minimum at x0 G (a, b). If / is differentiable at x0, then /'(*>) = 0. Proof Assume / has a relative maximum at x0. Then there is 8 > 0 such that, if x0 — 8 < x < x0 + S, then x G [a, b] [since x0 G (a, &)] and /(*) ^ f(x0). Consider any sequence {xn}^x converging to x0 such that x0 — 8 < xn < x0. Then, since / is differentiable at JC0, t

X

n ~ *0 Jw=i

converges to f'(xo). But f(x„) - f(x0) ^ for each n because f(xn) ^ f(x0) and xn < JC0, and hence /'(-^o) — 0* Consider now a sequence {^„}*=i such that x0 < yn < x0 + 8. As before,

f/(^)-/fa)r

1

yn - x0 / „ „ ,

converges to /'(Jt0), but this time f(yn) ~ f(xo) ^ 0 . yn - *o hence, f'(x0) < 0. Therefore, f(x0) = 0. The case in which / has a relative minimum at x0 is left to the reader. The reader should now realize that an important fact in this proof is that we are free to choose points in the domain on either side of x0 as close as we please. Since both the maximum and the minimum occurred at end points in the preceding example, the theorem did not apply. The following theorem, known as Rolle's Theorem, is an application of Theorem 4.6. We shall postpone a discussion of its geometric interpretation until we have completed the proof. 4.7 ROLLERS THEOREM Suppose / : [a, b] -» R is continuous on [a, b] and / is differentiable on (a, b). Then if f(a) = f(b) = 0, there is c G (a, b) such that f(c) = 0. Proof If f(x) = 0 for all x G [a, b], then f(x) = 0 for all x G [a, b\ and the theorem is proved. Suppose f(x) i1 0 for some x G [a, b]. By Theorem 3.7, [a, b] is compact since it is closed and bounded; hence, by Corollary 3.11, / assumes its maximum

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43 RoUe's Theorem and the Mean-Value Theorem 121 /'(c) = 0

Figure 4.2

and minimum in [a, b], say at xx and x2i respectively. Since / is not identically zero on [a, b] and f(a) = f(b) = 0, at least one of xx and x2 must belong to (a, b\ say xx G (a, b). Now, by Theorem 4.6, /'(*i) = 0. In essence, the theorem states that if the graph of a differentiable function touches the x axis at a and at b, then somewhere between a and b there is a horizontal tangent. See Figure 4.2. Note that the theorem states that there is at least one c such that /'(e) = 0. There may be more such points as indicated in Figure 4.2. If we examine the graph in Figure 4.2 from a geometric point of view—that is, ignoring the choice of coordinate system—it would appear that the tangent line at c is parallel to the line connecting the endpoints of the curve. This is essentially the content of the Mean-Value Theorem to follow. Suppose we have a smooth curve, the graph of /, connecting the points (a, f(a)) and (b, f(b)). We can rotate and translate the coordinate axes so that the points lie on the new x axis. Then RoUe's Theorem will guarantee a point on the curve where the tangent line is parallel to the new x axis, which contains the segment connecting the points (a, f(a)) and (b, f(b)). (See Figure 4.3.) The slope of the segment joining (a, f(a)) and (b, f(b)) is Kb) ~ f(a) so this is the slope of the tangent line at c; that is, b- a The precise statement and proof follow. 4.8 MEAN-VALUE THEOREM If f:[a,b]->R is continuous on [a, b] and differentiable on (a, 6), then there is a c E (a, b) such that

fie) =

Kb) - f(a) b — a

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122 Chapter 4 Differentiation

(M«)

Figure 4.3

Proof To prove this theorem, we shall find a linear function L such that f - L satisfies Rolle's Theorem and shall then apply Rolle's Theorem to obtain the desired result. This corresponds to the change of coordinate system mentioned before this theorem. The function L must be linear and satisfy L(a) = fid)

and

Lib) = f(b).

This is accomplished by choosing L(x) =

[fib) - f(a)1

(x - c) + /(a),

the equation of the line passing through (a, /(a)) and (b, fib)). L is continuous and differentiable everywhere and L'ix) =

fib) ~ fid) b- a '

Let g = / - L. The function g satisfies the hypotheses of Rolle's Theorem; hence, there is c S (a, b) such that 0 = g'ic) = f'ic) - Vic) = f'ic) -

fib) - fja) b- a '

Thus

f'ic) =

f(P) - fja) b- a '

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4.3 Rolle's Theorem and the Mean-Value Theorem 123

The Mean-Value Theorem has some very important applications in relating the behavior of / and /'. The next theorem will illustrate this. 4.9 THEOREM Suppose / is continuous on [a, b] and differentiable on (a, b). Under these hypotheses: i. If /'(*) * 0 for all x G (a, b\ then / is 1-1. ii. If /'(JC) = 0 for all x G (a, b), then / is constant. iii. If f'(x) > 0 for all x G (a, b), then x < y and x, y G [a, b] imply /(•*) < /(>) (that is, / is strictly increasing). iv. If f(x) < 0 for all x G (a, b), then x < y and x, y G [0, ib] imply /(*) > f(y) (that is, / is strictly decreasing). Proof Consider any JC, y G [a, &] with x < y. Now / is continuous on [JC, y] and differentiable on (JC, y); so, by the Mean-Value Theorem, there is c G (x, y) such that * - y

that is, /(JC) — /(y) = f(c)(x — 7). With this fact in mind, we shall proceed. (i) Suppose / is not 1-1. Then there are x, y G [a, b] with x < y such that /(*) = /O0« Thus» there is c G (JC, y) such that

= / - / w

m

x - y

=0

contrary to /'(c) # 0 for all c G [a, b]. (ii) Suppose / is not constant on [a, b]. Then there are x, y G [a, 6], JC < y, such that f(x) ¥= /(y). There is c G (a,fc)such that m

=

/ - /(y) x - y

#

0

contrary to /'(c) = 0 for all c G [a, &]. (iii) Suppose x < y and JC, y G [a, b). There is c G (JC, y) such that

m-m

= f (c) > 0.

hence f(x) < f(y). (iv) Suppose x < y and JC, y £ [a, b]. There is c £ (x, y) such that • y

hence /(x)

/'(c) < 0;

> /o».

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124 Chapter 4 Differentiation

Now that the proof has been presented, it is clear that items (iii) and (iv) can be changed as follows to give additional information: iii. fix) ^ 0 for all x E (a, b) implies / is increasing. iv. f(x) ^ 0 for all x E (a, b) implies / is decreasing. The following theorem is an obvious corollary to Theorem 4.9. 4.10 THEOREM Suppose that / and g are continuous on [a, b] and differentiable on (a, b) and that f{x) = g'(x) for all x E (a, b). Then there is a real number k such that f(x) = g(x) + k for all x E [a, b]. Proof Consider h(x) = f(x) - g(x) for all x E [a, b]. Clearly, h is continuous on [a, b] and differentiable on (a,fe)and *'(*) = /'(*) - *'(*) = 0 for all x E (a, 6). Hence, by Theorem 4.9, there is a real number k such that k = h{x) = f{x) - g(x) for all JC E [a, &]. Thus, /(*) = g(x) + jfc for all x E [a, 6], • Example 4.7 The Mean-Value Theorem has a variety of uses, one of which is that of estimating values of certain functions. Suppose p > 1 and h > 0. Define the function f(x) = (1 + x)p. Assuming the basic facts about the differentiation of powers, / is differentiable for x > 0. So, if h > 0, there is t such that 0 < t < h and

m =m 3 W

~m

A-0

/(A) = /'(*)(* ~ 0) + /(0) = Ml + 0*-1* + 1. Nowp > 1; hence, (1 + t)p~x > 1, so f{h) = (1 + /0P = p(l + 0P_1A + 1 > ph + 1. See Exercise 24 for the case of p < 1.

•

Consider a function / differentiable on [a, b]. If f(x) # 0 for all x E (a, 6), then by Theorem 4.9 / is 1-1 and by Theorem 3.19 / is monotone. If / is increasing, then, for all x, y E [a, b] with x # y, we have /(*) - / ( y ) ^ Q

x- y

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4.3 Rolle's Theorem and the Mean-Value Theorem 125 Hence, /'(JC) > 0 for all x E [a, ft]. Similarly, if / is decreasing, f'(x) < 0 for all x E [a, ft]. We have thus proved the following theorem. LEMMA

If / is differentiable on [a, b] and

/'(JC)

± 0 for all x E (a, ft), then

either /'(*) s> 0 for all JC E [a, 6] or /'(JC) < 0 for all JC E [a, 6].

Let us consider this result from a slightly different point of view. Suppose / is differentiable on [a, ft], and there are x, y E [a, 6] such that f(x) > 0 and f'(y) < 0. If we restrict our attention to [JC, y] (assuming JC < y), then, by the lemma, there is c E (JC, y\ such that /'(c) = 0. Thus, we are led to suspect that / ' has the intermediatevalue property. That this is indeed the case is stated in the following theorem. 4.11 THEOREM Suppose / is differentiable on [a, b] and A is a real number such that f(a) < A < f{b) or /'(ft) < A < /'(a). Then there is c E (a, b) such that/'(c) = A. Proof

Define

g(x) = fix) -

AJC

for all JC E [a, 6]. Then g is differentiable on [a, ft] and g'(jc) = /'(JC) - A. If /'(a) < A < /'(ft), then g'(a) < 0 and g'(b) > 0; if/'(ft) < A < /'(a), then $'(*) > 0 and g\b) < 0. In either case, g' has opposite signs at a and at ft. Hence, by the lemma, there is c E (a, ft) such that 0 = g'(c) = /'(c) - A so that /'(c) = A. Suppose / : [a,ft]-» /? is such that there is a differentiable function g : [a, ft] -* /? such that gfix) = /(x) for all x E [a, ft]. Now / need not be a continuous function, but / must have the intennediate-value property; in other words, if JC, y E [a\ ft] and Six) < A < /(y), then there is c between JC and y such that /(c) = A. This means that a function that is a derivative of some function must be special. In particular, Six) = [x] cannot be the derivative of a function. The following theorem is a generalization of the Mean-Value Theorem. 4.12 CAUCHY MEAN-VALUE THEOREM If / and g are continuous on [a, ft] and differentiable on (0, ft), then there is c E (a, ft) such that [/(» - /(a)]*'(c) = fe(6) - s(*)]/'(c). Proof Define *(*) = [/(ft) - f(a)]g(t) - [gib) - g{a)]f(t) for each r E [a, ft]. Note that A is continuous on [a, ft] and differentiable on (a, ft) and that hia) = fc(ft). So, by the Mean-Value Theorem, there is c E (a, ft) such that h'ic) = [/(ft) - Sia)]g'ic) - [g(ft) - g(a)]/'(c) = 0.

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126 Chapter 4 Differentiation

4.4 UHOSPITAUS RULE AND THE INVERSE-FUNCTION THEOREM We can use Cauchy's Mean-Value Theorem to prove a familiar rule from calculus, L'HospitaFs Rule. 4.13 UHOSPITAUS RULE Suppose / and g are continuous on [a, b] and differentiate on (a, b). If xQ E [a, b], i. g'(x) ± 0 for all x G (a, b\ ii. f{xQ) = g(xQ) = 0; and iii. f'/g' has a limit at xQ, then fig has a limit at JC0 and lim - (JC) = lim ~- (r). *-*o g

A-KV0 g

Proof Let {JC„},T=i be any sequence converging to x0 with *„ G (0, b)\{x0] for all «. By the Cauchy Mean-Value Theorem, there is a sequence {cw}~=1 such that (i) cn is between xn and * 0 for each n, and (ii) [/(x„) - f(xQ)}g'(cn) = [g(*„) - g(x0)]f(cn) for each a. Since xn ± x0 and g'(x) # 0 for all JC E (a, £), g is 1-1, hence g(xn) # g(jc0) = 0 for all n. Since f(x0) = g(x0) = 0,

f(xn)

=

g(jcw)

/(*„) ~ /to) = /'(c) g(*„) - g(jcb)

g ' W

and since / ' / # ' has a limit at x0, the sequence

converges; hence, so does the sequence

I7(**)l and both to the same limit. Therefore, fig has a limit at x0 and lim ( £ ) (*) = lim ( S - J (JC). • Example 4.8 To illustrate the use of L'Hospital's Rule, consider the functions / and g defined as follows:

f(x) = Vx - V2 + VJC - 2

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4.4 L Hospital' 5 R ule and the Inverse-Function Theorem 111 and g(x) = V.v2 - 4,

for all x G [2, 3].

Now / and g are both continuous on [2, 3] and differentiable on (2, 3), and /(2) = g(2) = 0. Further, f'(.v) = —7= + — - =

and

/>'(.v) = - = = = .

So £Cv)

__ (V.v

-

S'(A)

2 4- VAQV.Y -

2V.v

+ 2

(2VvV.v - 2)(A) ~"

2.vVv

*

1

It is clear that fig has a limit at 2 and that limit is 1/2. So, by L'HospitaPs Rule, fig has a limit at 2 and the limit is 1/2. • The next theorem is a one-dimensional version of the Inverse-Function Theorem. In Chapter 3, we observed that, if / : [a, b\ -* R is continuous and 1-1, then /'"' is also continuous. It is now natural to inquire what can be obtained if it is assumed that / is differentiable. Before proceeding further, let's look at the possibilities. If / is 1-1 and differentiable, and i f / ' ' is differentiable, then, by the Chain Rule, CT 1 •/>'(*) = (/' 1 ) , (/(.v))/'(.v), but (/"' o /)(.V) =

x

for all .v G \a, h}\ hence,

1 = < r , 0 / ) ' ( . v ) =

1 2

1+*2'

Of course, this is the usual formula for the derivative of Arctan x.

•

EXERCISES 4.1 THE DERIVATIVE OF A FUNCTION 1. Let (JC0, yo) he an arbitrary point on the graph of the function f(x) = JC2. For x0 ¥= 0,findthe equation of the line tangent to the graph of / at that point byfindinga line that intersects the curve in exactly one point. Do not use the derivative tofindthis line. 2. Prove that the definition of the derivative and the alternate definition of the derivative are equivalent. 3. Use the definition tofindthe derivative of f(x) = VJC, for x > 0. Is / differentiable at zero? Explain. 4. Use the definition tofindthe derivative of g(x) = x2. 5. Define h(x) = x3 sin \ for x # 0 and h(0) = 0. Show that h is differentiable everywhere and that h' is continuous everywhere but fails to have a derivative at one point. You may use the rules for differentiating products, sums, and quotients of elementary functions that you learned in calculus. 6. Suppose / : (0, 6) -» R is differentiable at x G (a, b). Prove that r /(* + *) - /(* ~ *> urn — exists and equals /'(JC). Give an example of a function where this limit exists, but the function is not differentiable. 7. A function f:(afb)-*R satisfies a Lipschitz condition at JC G (a, b) iff there is M > 0 and € > 0 such that |* - y| < e and y € (a, 6) imply that |/(JC) - f(y)\ < M|JC - y\. Give an

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130 Chapter 4 Differentiation example of a function that fails to satisfy a Lipschitz condition at a point of continuity. If/ is differentiable at oc. prove that / satisfies a Lipschitz condition at .r. 8. A function / : (a, b)—> R is said to be uniformly differentiable iff/ is differentiable on (a, h) and for each € > 0, there is 8 > 0 such that 0 < |.\- - v| < 8 and x, y G (a, b) imply that

K^-f(.v)|