Mathematical Analysis in Questions and Problems 5030002170

Table of contents :
Front Cover
Title Page
Contents
Preface
List of Symbols
1. Real Numbers
2. Limit of a Sequence
3. Limit of a Function. Continuity of a Function
4. Derivatives and Differentials
5. The Indefinite Integral
6. Fundamental Theorems on Continuous and Differentiable Functions
7. Investigating the Behaviour of a Function and Constructing Graphs
8. The Definite Integral
9. Lebesgue Measure and Lebesgue Integral
ANSWERS AND HINTS
Index
Back Cover

Citation preview

MATHEMATICAL ANALYSIS in Q uestions and Problems Mir Publishers Moscow

MATHEMATICAL ANALYSIS IN QUESTIONS AND PROBLEMS

M aieM aT H n ecK H H a n a j if i3 b Bonpocax h 3aja;aHax IIoA peflaKijneH B. O. ByTyaoBa tlsfluTejibCTBo «Bi>icman mKOJia» Mockhu

MATHEMATICAL ANALYSIS in Q uestions and Problems Under the Editorship of B.F.Butuzov

Mir Publishers Moscow

Contributors: B. F. Butuzov, N. Ch. Krutitsknyn, G. N. Medvedev, A. A. Shishkin

Translated from the Russian by Irene Aleksanova

First published 1988 Revised from the 1984 Russian edition

Ha aMAuucKOM jt.iuxe

Printed in the Union of Soviet Socialist Republics

ISB N 5 -0 3 -0 0 0 2 1 7 -0

© H3AaTejn>CTBO «BbicmaH niKOJia», 1984 © English translation, Mir P u b lish ers, 1988

Contents

P reface

7

List of Symbols

oo 7 1 -0 0

lim / (.r) - ft, / (*) 6 as ,r

a

The upper limit of the sequence (crn) The number b is a limit of the function / (x) as x tends to a

lira / (a:) *-»a+0 lim f (x) x-+a—0 lim / (x) x-**a+0 lim / (x) x-+a-0 inf / (x) x sup / (x)

= = = =

b, b oo, oo

sgn x a ~ (3 as x —wt a = o (P) as x—a A!/ /' (^). j/' (*) /' (x -I- 0), r {x - u) dy /(x) dny r = r (0 J / (*) dx h 1 1 (x) dx a j / (*) dfi (r) f (x) = g (x) on £

The number h is a limit of the function / (x) as x tends to a from the right (from the left) The function / (x) is infinitely large on the right (on the left) of the point a The greatest lower bound of the function / (x) on the set X The least upper bound of the func­ tion / (x) on the set X The Dirichlet function The integral part of the number x The signum function The infinitesimals a and (1 are equivalent as x -*-a a (x) is an infinitesimal of a higher order of smallness than (^) x as x —>■a The increment of a function at a point Ihe derivative of the function y = / (x) at the point x The right-hand (left-hand) deriva­ tive at the point x The differential of the function y (x) The nth derivative of the function / (*) The nth-order differential of the function y (x) A vector function Ihe indefinite integral of the func­ tion / (x) Ihe definite integral of the function / (x) over the interval [a, b] The Lebesgue integral of the func­ tion / (x) over the set E The functions / (x) and g (x) are equivalent on the measurable set E

Chapter 1

Real Numbers

1.1. Comparison of Real Numbers I. Fundamental Concepts and Theorems 1. Representing real numbers as nonterminating deci­ mal fractions. Any real number a can be represented as a nonterminating decimal fraction CL = -f-flQ.tZjfln.. Mn... T where we choose one of the two signs plus for pos­ itive numbers and minus for negative numbers (the plus sign is usually omitted). Rational numbers can be represented as repeating and irrational numbers as nonrepeating nonterminating deci­ mal fractions. Some rational numbers can be represented as a terminating fraction or, what is the same, as a non­ terminating fraction with zero in the period. Numbers of this hind admit of another representation, as a nonter­ minating decimal fraction with the figure nine in the period. For instance, 1/2 = 0.500...0... - 0.50, 1/2 = 0.4999...9... = 0.49. When comparing real numbers, we shall use only the lirst form of notation for rational numbers of this hind (with zero in the period). 2. Rule of comparison of real numbers. Let a = and 5 = ± b Q.b1b2...bn... be arbitrary real numbers represented as nonlerminating decimal fractions. The numbers a and b are equal (a = b) if they are of iJie same sign and the equalities ah = bh (k = 0, 1, 2 . .) hold true. Otherwise we assume that a. =£. b. ' When comparing unequal numbers a and b, we shall consider three cases: (1) a and b are nonnegative numbers. Since a ^ b, tlieye is a natural number n (or n = 0) such that ah s= bh

(k = 0, 1, . . ., n — 1) and an =^= bn. We assume that a > b if au > 6n ancl a > b if | a | < | b | and a < b if | a | ;> | b |. 3. Some number sets. Real numbers can be given as points on a coordinate line*. Therefore the set of all real numbers is called a number line (number axis) and the numbers are called points, and the geometric interpreta­ tion is often used when number sets are considered. We shall use the following designations and terminology: N is the set of all natural numbers, Z is the set of all integers, R = (—oo, -f-oo) is the set of all real numbers (the number line), [a, b] is a closed interval or the set of all real numbers x which satisfy the inequalities a ^ x ^ b, (a, b) is an open interval or the set of all real numbers x which satisfy the inequalities a < x < b, [a, b) or (a, fc] is a half-open or half-closed interval oi the set of all real numbers x which satisfy the inequali­ ties a ^ x < b, a < x ^ b respectively, [a, -f°°) or (a, + oo) or (—oo, a] or (—oo, a) is a naif-Line or the set of all real numbers x which satisfy the inequalities a ^ x < -|-oo, a < x < -|-oo, —oo < x ^ a, —oo 0. II. Control Questions and Assignments 1. What is the difference between the lionterminating decimal fractions which represent rational and irrational numbers? 2. In what case are two numbers equal? 3. Do the equalities 0.419 = 0.420 = 0.42 hold true? 4. Formulate the rule of comparison of two unequal numbers. * Recall that the coodinate line is a straight line on which an arbitrary reference point, called the origin, a unit of length and a positive direction are chosen.

12

III. Worked Problems 1. Prove llial for any real numbers a and b (a flj, and smaller than b since ch = a.h = bh (k = 0, 1, . . n — 1), cn = an < bn. Thus there is a rational number c such that a < c b and b >- c, then a > c. 7. Prove that the inequalities —| a \ ^ a < | a | bold true for any number a. 8. Prove that if x < y, then —x > —y. 1.2. The Least Upper and the Greatest Lower Bound of a Set. The Use of the Symbols of Mathematical Logic I. Fundamental Concepts and Theorems 1. On the use of some logical symbols. Let X be a non­ empty set of real numbers. Definition. The set X is bounded front above (from below) if there is a number M (m) such that for any number x from the set X the inequality x ^ M (x ^ m) holds true. The number M (m) is the upper (the lower) bound of the set X. ^ ; In this definition and in the formulations of many other definitions and theorems use is made of the words “there is and for any”. To make the notation shorter, we shall use the logical symbols 3 and V instead. The symbol 3 is the existential quantifier and the sym­ bol V is the universal quantifier. The fact that the number x belongs (does not belong) to the set X will be designated a s x e x (z$ x ). Using these symbols, we can write the definition of a set bounded from above as follows: the set X is bounded from above if 3M 6 R such that \fx 6 X the inequality x ^ M holds true, or (even shorter, omitting certain words), the set X is bounded from above if 3M 6 R Vx e X: x < M. (1) The use of quantifiers not only makes the notation short­ er bill also allow s US 10 construct negations of propositions 14

(definitions, assertions) in a simple way. W e shall illu­ strate this technique by an example of the negation of the definition of a set bounded from above. In other words, we shall formulate the definition of a set unbounded from above. The unboundedness of the set X from above means that there is no number M such that for any x £ X the inequality x ^ M holds true. This means that for any number M there is x 6 X for which x > M. Using Lhe quantifiers, we can write the definition of a set un­ bounded from above as follows: the set X is unbounded from above if ViW 6 R 6 X: x > M. (2) Comparing (1) and (2), we see that to construct the negation of proposition (1), we must replace the quantifier 3 by V and the quantifier V by 3 and replace the inequality appearing after the colon sign by exactly the converse proposition. This rule can also be used to construct negations of other statements containing the quantifiers 3 and V. 2. The least upper bound and the greatest lower bound of number sets _ Definition. The number x is the least upper bound of the set X bounded from above if (1°) Vx 6 X: x ^ x, (2°) Vx < x 3x 6 X: x > x._ Condition (1°) means that x is one of the upper bounds of the set X and condition (2°) means that x is the least of the upper bounds of the set X, i.e. no number x, which is smaller than x, is an upper bound. The least upper bound of the set X is designated as sup X. The greatest lower bound* of the set X bounded from below is defined by analogy and designated as inf X. Theorem. A nonempty set bounded from above (from be­ low) has the least upper (the greatest lower) bound. If the set X is not bounded from above (from below), then we write sup X = -foo (inf X = —oo). The set X is bounded if it is bounded both from above and from below, i.e. 3M, raVx 6 X: m < x < M. (3) * You may come across books on analysis where the least upper bound (the greatest lower bound) is simply called the uppor

(the lower) bound. 15

II. Control Questions and Assignments 1. Use quantifiers to write a definition of a set bounded from below. Construct the negation of this definition using the rule of constructing negations. 2. Give a delinition of the least upper (the greatest lower) bound of a set bounded from above (from below). 3. Formulate the theorem on the existence of the least upper bound and the greatest lower bound of a number set. 4. Prove that the least upper bound and the greatest lower bound are unique, i.e. that a set bounded from above (from below) has only one least upper bound and one greatest lower bound. 5. Show that the least upper bound and the greatest lower bound may or may not belong to a set. Give exam­ ples of number sets X in which (a) sup X £ X, (b) sup X $ X, (c) inf X £ X, (d) inf X $ X. Does the number X have the greatest number in cases (a) and (b) and the least number in cases (c) and (d)? 6. What is the meaning of the symbolic notation (a) sup X = + oo, (b) inf X = —oo? 7. What set is said to be bounded? 8. Prove that the following definition of a bounded set is equivalent to that given in subsection I: the set X is bounded if 3A > 0 Vx £ X: \ x | < A. Applying the rule of constructing negations to the definition given in assignment 8, formulate the delinition of an unbounded set. III. Worked Problems I. Find the least upper bound of the interval (0, 1). A The number 1 is an upper bound of the interval (0, 1) since Vx £^0, 1): x < 1. Furthermore, Vx < 1 3a £ (0, 1): a > x. Indeed, if x < 0, then Va £ (0, 1): a > x. If x > 0, then, as was shown in Example 1 in 1.1, there is a rational number a on the interval (x, 1) such that x < a < 1, i.e. 3a £ (0, 1): a > x. Thus, both condi­ tions for the definition of llie least upper bound are ful­ filled for the number 1. Consequently, sup (0, 1) = 1. Note that the least upper bound we have found does not belong to the interval (0, 1), i.e. sup (0, 1) $ (0, 1), 16

whereas lor the interval (0, 1] we have sup (0, 1] = 1 £

(0, 11 A

2. Find the least upper bound and the greatest lower bound of the set of all proper, rational fractions min (.m, n £ N, m < n) and show that this set does not have either the least or the greatest element. A Lot X be the set of all proper rational fractions rn/n. Since Vm, n £ N min > 0, the number 0 is a lower bound of the set X. Furthermore, Vx > 0 3« 6 X: a < x.

(4)

/V

Indeed, if x ^ 1, then the proper rational fraction a = 1/2 satisfies condition (4). If 0 < x < 1, then the number x can be written as a nonterminating decimal fraction /V

x = 0.x1x2...xtl...i 3n being such that xn ^ 0. According to the rule of comparison of real numbers, the rational number a = 0.a;1a:;,...a;n_1 (xn — 1)1 satisfies the inequalities 0 < a < x «< 1, i.e. is a proper rational fraction and satisfies condition (4). Thus the second condition of the definition of the great­ est lower bound of a number set is fulfilled for the num­ ber 0, and so inf X = 0. Since the set X contains only proper fractions, i.e. m < n, we have min < 1. Hence the number 1 is an upper bound of the set X. Furthermore, Va: < 1 3min 6 X: min >» x. Indeed, as was shown in Example 1 in 1.1, there is a rational number xx such that £ < < 1Since x1 < 1, it follows that x± is a proper fraction: xx = rain (m < /z), i.e. 6 X . Consequently, both conditions for the definition of the least upper bound of a number set are fulfilled for the number 1. Thus sup X = 1. However, inf X = 0 $ X since rain = 0 only for 7 7 i = 0, but 0 (£ N. This means that the set X does not have the least element. Precisely for the same reason sup X = 1 £ X since m!n = 1 only for ra = n, and this 2-Q1692

17

contradicts the requirement that the fraction should be proper. Hence the set X does not have the greatest ele­ ment. ▲ IV. Problems and Exercises for Independent Work 9. Assume that X and Y he nonempty sets ol' real numbers, X is hounded from above and Y is contained in X. Prove that Y is also hounded from above and sup Y ^ sup X. 10. Find the least upper bound and the greatest lower bound of the set of rational numbers x which satisfy the inequality x2 x . Since xx ^ x and yt ^ y, it follows that (xx + y^)T ^ (x + y)T (for rational numbers this property of inequali­ ties is known). Consequently, condition 1° is fulfilled. Let us show that condition 2° is also fulfilled. Let x be an arbitrary number smaller than (x + y)T. Between the numbers x and (x -)- y)r there is a rational number a (see Example 1 in 1.1) such that x < a < (x + V)r• We set 8 = (x -(- y)T — a (the subtraction is carried out according to the rule of subtraction of rational numbers). Then a = (x -f y)T — 8, and since 6 > 0, there is a natural number n such that 2In < 6. We shall consider now rational numbers x, = x — — 1

‘

n

and yx— y —. Since x , < x and yt < .y, it follows that (xj -j- y!)r ^{(x, + i/t)r}, and then (xx + z/,)r = l/)r ^ " > ( a: + y)r —8 = a since — n a > x, i.e. (xx + z/x)r > x. We have thus shown that condition 2° is fulfilled. A 2. Prove that Vx: x + (—x) = 0 . Let xx and yx be any rational numbers which satisfy the inequalities xx ^ x, yx ^ —x. We must prove that sup {(xx -f yx)r} = 0, i.e. (1°) V(xx + y x) T 6 {(*1 + y 1)r}- (*, + yx) r < 0, A/ (2°) Vx < 0 3 (xx + yt)T 6 {(*, + ?/i)r): (x, + yx)r > X. Since yx ^ —x, it follows that —i/x x (it is easy to establish this fact using the rule of comparison of real numbers, see Exercise 8 in 1.1). By virtue of transitivity of the sign it follows from the inequalities xx ^ x and x ^ —^x that xx ^ —yx and, hence, (xx -f- yx)T ^ 0. Thus condition 1° is satisfied. 20

We shall show that condition 2° is also satisfied. Let x be an arbitrary negative number. Since —x > 0, there is a natural n such that l/10n < — i.e. —1/10" > x. Let us represent the number i as a nonterminating deci­ mal fraction (for definiteness, we assume that a: > 0): x = x0.x1x.z...xn... . It follows from the rule of comparison of real numbers that Xj — Xg.XiX.2- •5^ 1 ^ h! — Xg.x^Xn. • .•^n ^ x. Thus fa + r)iJ6{fa + J/i)r}i and then fa -f —1/10" > x,i.e. we have proved that condition 2° is fulfilled. A IV. Problems and Exercises for Independent Work 12. Prove that multiplication of two rational numbers according to the rule of multiplication of real numbers yields the same result as multiplication according to rule (1) of rational numbers. 13. Prove that Vx: x -f- 0 = x. 14. Prove that V.r, y there is a unique number z such that x = y + z (z is the difference of the numbers x and z/, i.e. z = x — y). 15. Prove that Vx: x>\ = x. 16. Prove that V# 0 3x': xx = 1. 17. Prove that Vz and V?/ 0 there is a unique num­ ber z such that x = yz (z is the quotient of the division of x by y, i.e. z = x/y). 18. Prove that Va;, y, z: (x y) z = xz -|- yz. 19. Prove that if x > y, then Vz: x-f- z > y + 20. Prove that if x > y,then Vz > 0 : xz > yz. 21. Prove the validity of the following inequalities (a) I x + y | < | x | + | y |, (b) | x —y I > I x \ — \y\, 22. Assume that X and Y are nonempty bounded sets of real numbers and T is the set oT various sums x + y, where x £ X and y 6 Y. Provo that the set T is bounded and llml (a) sup T = sup X -I- sup Y , (b) inf T = jnf X + i»f Y, 21

=

23. Assume that X and Y are nonempty bounded sets of nonnegative real numbers and B is tbe set of various products xy, where x £ X and y £ Y. Prove that the set B is bounded and that (a) sup B = sup X -sup Y, (b) inf B = infX *inf Y. 24. Calculate the first three significant digits of the

sums (a) -1 + 1/3, (b) 1/3 + 1/7. 25. Find the first three decimal digits of the pro­ ducts (a) 4 V"3, (b) 1/3• l/7 . 26. Assume that A and B are nonempty sets of real numbers in which every number from A is smaller than any number from B, and for every e > 0 there are x £ A and y £ B such that y — x < e. Prove that sup A = inf B.

1.4. Induction I. Fundamental Concepts lo prove that a statement is true for any natural num­ ber n beginning with n0, it is sufficient to prove that (a) the statement is true for n = n0, (b) if the statement is true for some natural number k > n 0, it is also true for the next natural number k + 1. This method of proof is known as m athem atical induc­ tion.

II. Control Questions 1. What is the gist of the method of induction? 2. Use induction to prove that V n fN : III. Worked Problems 1. Prove that V« £ N and V.r > —1 the inequality (1 + x)n > 1 + nx (Bernoulli’s inequality) (l) holds (rue. A We shall use induction to prove inequality (1). If n — 1, then inequality (1) holds true since it becomes a (rue equalily. We assume lhat relation (1) holds (rue for a natural number k and V.r > —1: (2) 22

Since x > —1, it follows that 1 + %> 0. We multiply inequality (2) by the positive number +1 + -r: (1 + x)h+1 > 1 + kx + x + kx2. Removing the nonnegative term kx- on the right-hand side, wc obtain an inequality

(1 + x)h+1 > 1 + (k + 1) x. We have proved that inequality (1) is valid for the natui’al number k -f 1 and Vo; > —1. We have thus proved that relation (1) is valid Vn 6 N and Va: > —1. A 2. Prove that for any n positive numbers yu y2, . . (/„ which satisfy the condition VlVs — Vn = Ir

(3)

there holds a relation lJ \ + U* +

• • • + Vn ^ n •

(4)

A It follows from condition (3) for n = 1 that yx = 1. Therefore relation (4) is satisfied. Assume that relation (4) follows from condition (3) for n = k and that k -f 1 positive numbers yv y2, . . ., yh, yn+\ satisfy condition (3). We shall prove that relation (4) is satisfied for them. If all these numbers are equal to unity, then their sum is equal to k -f- 1 and relation (4) holds true. Now if there is at least one num­ ber different from unity among the indicated numbers, then there is necessarily one more number which is not equal to unity, and if one number is larger than 1, then the other is smaller than 1. Without violating the gen­ erality, we assume that yh > 1 and yh+x 100: Thus a

o

\xnioo — a \ ^ e

r

;Y=1. for N — 2, for N = 100.

lira x n if there are e > 0 and a sequence of

numbers {«*} °{N = 1, 2, 3, . . .) such that nN > N and | x nN — a | ^ e. The geometric interpretation of this definition is: a -=f=. lim x n if there is a e-neighbourliood of the point 71 —►OO a outside of which there are infinitely many terms of the sequence. A n 2. Prove that the sequence x n = 2', 0 , 3rc 6 N for which | ,r„ | > M. We take an arbitrary M > 0 and any even number n which satisfies the inequality n > log2 M. For such an n we have xn = 2'* > 2,ob*m = M, and this is what we wished to prove. A 3. Using the definition of a limit of a sequence, prove that lim -^5 *3n —^- = 5. N - * CO

**

^

A We specify an arbitrary e > 0 and consider the modulus of the difference of the /ith term of the sequence and the number 5: 5-371 3 « -2

r 10 ’’ — 3” — 2 •

In accordance with the definition of a limit of a sequence, we must indicate a number N such that Vn > N the 27

inequality a £ r< e

(1)

is satisfied. To find the number N, we solve inequality (1) for n. We obtain n > lo g 3

-f 2) .

(2)

It follows from inequality (2) that we can take as N the integral part of the number log3 + 2) : W= [ lo g ,( ^ - + 2 ) ] . Indeed, if « > iV , then « > [ lo g 3(-— + 2 )] + l > l o g 3 ( ^ - + 2) , i.e. inequality (2) holds true, and this means that V n > N the inequality (1) also holds true. (Note that for e > 10 we have N ~ [log3

-f 2) ] ^ 0

and

therefore inequality (1) is valid Vn£N. ) Thus, for an arbitrary e > 0 we have indicated the number iV- [log3 ( ^ - + 2) ] , such that Vn> JV the inequality

holds true. And this means, according to the definition of a limit of a sequence, that i5-3n c A lim I n U T ^ 5 A 7 l-> o o * ^

4. Using the definition of a limit of a sequence, prove that lim - * = 0. 7i— ► 0 0 y n\ A We specify an arbitrary e > 0. Wo must indicate a number N such that V/i > N the inequality U Yn\ < 28

p

(3)

holds Iruo. We shall not try to find the least number N, beginning with which inequality (3) holds true, but shall indicate a larger number uby excess” and solve a simpler inequality 2!n < e. (4) Since Mn £ N: ?i\ > n (n/A) (prove this), it follows that \fn £ N the inequality i/y « i < 2in (5) holds true and therefore inequality (3) is a consequence of inequality (4). Solving inequality (4) for n, we obtain n > 2It. (6) We set N = [2/e]. If n > N, then n ^ [2/e] + 1 >• 2/e, i.c. inequality (6) is satisfied and, consequently, inequ­ alities (4) and (3) are also satisfied. Thus Ve > 0 3N (N = [2/e]) such that Vrc > N: l/j^n! < e. We have I thus proved that lim — p= = 0. A V n The examples we have considered show the way to prove that lim x n = a using the definition of a limit of a sequence. It is necessary to form an expression | x n — a | and choose (if it is expedient) a sequence {y„} such that, first, Vra | x n — a | ^ yn and, second, for an arbitrary e it is easy to solve the inequality Vn < ® (7) for n. Assume that the solution of inequality (7) has the form n > / (e), where / (e) > 0. Then we can take [/ (e)] as N (if it turns out that [/ (e)] = 0, then inequality (7) is valid V n). Thus V« > N = [/ (e)I there holds an inequality I xn — a | < e and this means, according to the defini­ tion of a limit of a sequence, that lim xn = a. n-*-oo 5. It is known that lim xn = 0 and xn ^ 0 Vrt. n-*>oo Prove that lim x% = 0 for a > 0. 71— >00 A By the hypothesis, lim x n = 0, i.e. Vej > 0 3/Vi TL-+OQ such that V» > N j the inequality I *» I < 8i (8) 7 i —* 0 0

1

7 l-> 0 0

29

holds true. We have to prove that Ve > 0 3iV such that V/? > N: | x% | < e or, what is the same, l-c,,! < e1/a. (9 ) We specify an arbitrary e > 0 and set Ej = £*/“ (S] > 0). For this P] 3A^ such that Vn > iV, inequality (8) holds true, i.e.| a.„| < £ '/ “. Thus V/2 > N = inequality (9) holds true. We have thus proved that lim .t“ = 0. A IV. Problems and Exercises for Independent Work 1. Find out whether each of the following sequences is hounded: (a) x n = (—1)"—, (b) xn = 2n, (c) x n — In n, (d) xn = sin n, (e) {x„} = 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, . . . . Substantiate the answer. 2. Using the definition of a limit of a sequence, prove that (a) lim Tl—*-00

(~ l)n

n

= o,

(c) lim - °*n = 0, n-+oo

0 3TV such that Vrc > TV there holds an inequality | a |n > 4 . (1) We specify an arbitrary 4 > 0. To find the number TV, we solve inequality (1) for n. We get n > log,,,, 4*. (2) 7 !-> 0 O

We set N = [logia( 4]. Then Vra > N inequality (2) holds true, and, lienee, inequality (1) also holds true. Thus V4 > 0 3W = (log|a | 4] such that Vrc > N: | a |” > 4 . And this is what we had to prove. (b) Let | a \ < 1. If a = 0, then a” = 0 Vn and, consequently, {a"} is an infinitesimal sequence. Let a 0. Then an = ((1/a)")-1. Since | 1/a | > 1, the se­ quence {(l/a)'1} is infinitely large and the sequence ((l/a)n)-1 is infinitesimal by virtue of Theorem 5. Thus the sequence {«"} is infinitesimal for | a | 0 the term of the sequence with the number n = 2 ([71/] -f 1) is equal lo n and exceeds 71/, and this means, by definition, that {:rn} is an unbounded sequence. (b) We shall prove that the sequence {x-„} is not infi­ nitely large. Indeed, the interval (—2, 2) evidently 3 -0 1 G 3 2

33

includes all terms of the sequence with odd numbers and, hence, the interval contains infinitely many terms of the sequence {#„}. It follows that {a;,,} is not an infinitely large sequence. A 4. Assume that {.rn} is a convergent sequence and {i/n} is an infinitely large sequence. Prove that the sequence {^n + J/n) is infinitely large. A We shall prove that the sequence {;rn -|- i/n} satis­ fies the definition of an infinitely large sequence, i.e. VA > 0 3N such that >• N: | x n -|- yn | > A. Since {a:n} is convergent, it is bounded, i.e. 3M > 0 such that Vn there holds an inequality I

I< M .

(3)

We specify an arbitrary A > 0. Since {i/n} is an infinitely large sequence, it follows that for the number A + M 3 N such that Vrc >> N we have I Vn I > A + M.

(4)

From (3) and (4) we find that V« > N there holds an inequality \ X n + V n \ > \ V n \ — \ X n \ > A - \ - M — M = A, and this is what we had to prove. A IV. Problems and Exercises for Independent Work 7. It is known that in a neighbourhood of zero there are (a) a finite number of terms of a sequence, (b) an in­ finite number of terms of a sequence. Does it follow that in each of these cases the sequence is bounded? infinites­ imal? infinitely large? 8. It is known that the sequence {;rn} is convergent and the sequence {r/n} is infinitely large. Can the .se­ quence {xnyn} he (a) convergent, (b) divergent but bound­ ed, (c) infinitely large, (d) infinitesimal? Answer the questions using the sequences {«}, j 11~ 1 } , ( 4 "} ’ { (~ 1)n } > { ^ 5-} as examples. 9. Give examples of sequences {xn} and {yn} f°r which lim x n = 0, lim y n = oo and their product {x„yn} 71-VOO

34

TH-OO

is (a) a convergent sequence, (b) a divergent but bounded sequence, (c) an infinitesimal sequence, (e) an infinitely large sequence. 10. Prove that the following specified sequences are infinitesimal: (a) xn = nh (/c < 0), (b) xn = ( — l )11-0.999“, Xn = TH" ’ ^

Xn =

2»3+ l •

11. Prove that the following specified sequences are infinitely large: (a) x n = nk (k > 0), (b) x n = n (—1)”, (c) x n = 2V”, (d) x n = logo (logo n) (n > 2). 12. Prove that any infinitely large sequence is un­ bounded. 13. Prove that the sequence {(1 -f (—I)’1) n} is un­ bounded but not infinitely large. 14. Prove that if lim x n — +oo (—oo), then the 71 — *■OO sequence {a:,,} attains its greatest lower (least upper) bound. 15. Find the smallest term of the sequence {.?„} if (a) xn — n2—9n — 100, (b) xn = n-\- — .

2.3. Properties of Convergent Sequences I. Fundamental Concepts and Theorems Theorem

. Let lim.rn = a, lim yn — b. Then

6

71—>CO

77~►oo

(a) lim (.rn + yn) — a-\-b, 1 1 -+ 0 0

(h) lim (xnyn) — ah, 71—►OO

(c) if b =£ 0, then, beginning with a certain number a sequence {xn/yn} is defined (i.e■ 3lV such that V/? A1'. y n ^ 0) and lim (x nlyn) = 71->00

If lim xn — lim yn = 0, then lim (xn/y„) is an indeter71—►OO

71 -*00

71->00

minate form 0/0. The indeterminate forms oo/oo, 0-oo, oo — oo are defined by analogy. It is clear that Theorem 6 is inapplicable for these limits. Theorem 7. I f lim x n = a. and, beginning with a certain 71-►OO

number, x n > b (.r„ < b), then a ^ b 3*

(a ^b ). 35

Theorem

8

(theorem on three sequences), tf lim x n^ a, 71 -*■ oo

lim y n = a and' beginning with a certain number % the inequalities x n ^ zn ^ \jn hold true, then lim zfl — a. n -* o o

11— >

II. Control Questions and Assignments 1. Give .1 definition of a convergent sequence. 2. Use the “e-iV” language to give the definition of a divergent sequence and give a geometric interpretation of this definition. 3. Formulate Theorems 0-8. 4. Let the sequence {zn} converge and the sequence {yn} diverge. Prove that the sequence {x„ y n} di­ verges, the sequence { b Vrc. Does it follow 71-►oo that (a) a >> /;, (b) a ^ fc? III. Worked Problems 1. Assume that lim y n = b ^ = 0 and the sequence 71 -►OO

{zn} is divergent. Prove that {zny n} diverges. A We designate x n = zny n and prove the divergence of the sequence {r7,} by the method of indirect proof. We assume that {#„} converges. Since lim y n = b^= 0 71-►OO ]jy the hypothesis, it follows, according to Theorem 6, that Ihc sequence {;rn/j/n} = {z7,} is defined beginning with a certain number and is convergent. But this con36

tradicts the hypothesis. Consequently, is diver­ gent. A 2. Prove that the sequence {sin /?} is divergent. A We shall use the method of indirect proof. Let sin n = a. Then lim sin (n -f 2) = a, whence 7 t—* c o

n -> o o

lim (sin (n + 2) —sin n) —0.

(1)

ll-> o o

Since sin (;t + 2) — sin n = 2 sin 1 cos (» -f 1), we can take relation (1) into account and obtain lim cos (n + 1) = 0. (2) From the relation cos (n -f 1) = cos n cos 1 — sin n sin 1 we find thal sin n = s;n~j (cos n cos 1 — cos (n 1))From this it follows, by virtue of (2), that lim sin n = 0. 71-+-CO Thus we get lim cos n = lim sin n = 0, and Ibis conu-*-oo n -> oo tradicts the equality cos2 n + sin2 n = 1. Consequently, {sin «} diverges. A 3. Find the following limits: , , lim 2+1 10 (c) lim 5-3’1 (b) lim (a) *1_w

n— .oo n— Jr n

1

3"—2

A Note that each of these limits is an indeterminate form of type oo/oo. We have 10/1 !. 10 0 (a) n~*°° n+ — Ti =

since (fi)

*s an infinitely large sequence, n2—n («—\ rn) (n+ \rn) lim ----- Tr- = lim

„_»oo

n—V

»

n — l^n

71-.0O

= lim (n + yn ) = + 00» 5-3"

_

(c) lim ~gnZ-2 ~ ^ n

—

1

3n

■= 5. A lim U - — ) 71-*-OO' ° ' 37

4. Find the limit lim ({An24 -« — n). 7 1 -*

OO

A Note that this limit is an indeterminate form of type oo — oo. We have lim (Y n2+ n —n) — lim — - n ---71—voo

7 l-* o o

= lim ----- =====---n"~ l / 1+ - + 1

y

ft?

TL- f - n

_______ 1________ lim [ \ / 1 + — ) + l

„-oo \ y

nI

cos n n+1 A The sequence {cos n} is bounded and the sequence

5. Calculate lim

{

Yn

} 4S infinitesimal since lim

»-co

Yn

_J *- =

»+1

lim

/ n

lira 71—> OO

Yn ■=

+— n

1

lim

n-voo

0

.

(•+4-)

From this, according to Theorem 4, it follows that the product of these sequences is an infinitesimal sequence, i.e. lim — n 003” = Q a «->«, « + l A . Find the limit lim i 4+ 24+34+ ... + »4 7 1 -voo re6 A We designate S n = l 4 4- 24 4- 34 4- . . . + »4. We sJiall seek S n in the form S n = A n 5 + Bn 4 + Cra3 + Dn2 + En + F, and then Sn+1 - Sn = A l(n + l )5 - n5] + B [(« + l )4 _ 7l4! + C [(» + l )3 - «3].+ D \{n + l )2 _ «2] + E [(« -f 1) — n]. 6

From tin's we Iiave the following relation for any natural n: (n + l )4 = 5A n1 4- (10^4 4- 45) nz 4- (10^4 4- GB + 3C) n 2 4- (5A + 4B + 3C 4- 2D) n + A + B C + D 4- E. 38

Equating the coefficients in the same power of n on the left-hand and right-hand sides of the relation, we obtain 5A = 1, 104 + 4B = 4, 104 + 65 + 3C = 6, 5A + 4B + 3C + 2D =4, A + B + C + D + E = 1. Hence 4 = 1/5, 5 = 1 /2 , C = 1/3, 5 = 0, 5 = — 1/30. I 5-f-l-y nl4-f y 7i3— Thus, for any n, we have Sn = -g-re Setting n= 1, we get 1 = -g- + —■+ -gQ-

—

+ E, whence F = 0. Consequently,

S n = l* + 24+ 34+ • •. + n 4= — •+

.

OU

Thus we have l im 14+24+34+ - - - + ra4 Jl-voo W / 1 . 1 . 1 1 \ —i™

( 5 ^ 2n +

3n2" 3 0 ^ ) ~ T

1

' *

IV. Problems and Exercises for Independent Work 16. (a) It is known that the sequence {an} is convergent and {yn} is divergent. Can the sequence {xny n} be con­ vergent? divergent? (b) The sequences {#„} and {(/„} are known to be diver­ gent. Can the sequences {.t„ + yn} and {x„yn} be con­ vergent? divergent? Answer the questions using the sequences > {( —I)"}- {4 "}’

( —”}* {(— l)n+1} as examples.

17. Given the sequences j y j ’ {*4”} ’ {

}’

{~»-H 00 } ’ c^loose fr°m these infinitesimal sequences such that (a) lim (xnhjn) = 0, (b) lim (xjy„) = 1, n -* o o

? i—►oo

(c) lim (xn/yv) = oo, (d) {#„/#„} diverges but is bounded. 39

18. Given lim xn = b =f=oo, limi/„ = oo. Prove that 71-►OO

71—►OO

(a) lim (xn ± y„) = oo,

(b) lim (x jx jn) = 0,

71—> OO

71“ ►OO

(c) lim (yn/xn) =

00

foi =9^ 0),

(d)

lim (a^i/J = oo if

71—>00

U -> 00

ft =j4=0. 19. Prove that lim (xn/yn) = oo if lim xn = b=£ 0, 71—►oo 7 l“ >Oo lim i/n = 0 (i/„¥= 0). 71—►oo

20. Test the following sequences for convergence (depending on a, p, 7):

W-- S-’ (b> ^, ^ 5

1 7

21. Find the following limits:

(a) lim T/r^ T 2(”!) » (b) lim (Vn + 7l-> oo

, N 1.

1 ^

;•

—Vn),

1

71—►OO

( —2)n + 3n

(c) n-*oo lim (/ -_22)\n+i I qn+i • ™ + 3™ n

. Let xn = 2 ,/■■„ 1 • • Calculate lim a;,, 1 . 1 Vn k 7l— ► 0 0 /i=i evaluate xn from above and from below: 22

■ oo

|

——0 -f- 0

. 7i2 ~ h 2

. -}- 0 ;—0.

n 2 - |- n

We have thus found that 1 = 0. Find the error in the reasoning. Which of the two results is true and which is not, and why? 23. Calculate the following limits:

b(b+ 1)(b+ 2) )•

2.4. Remarkable Examples I. Fundamental Concepts and Theorems We say that the infinitely large sequence {#„} has a higher order of growth than the infinitely large sequence {l/n} if {xnfyn} is an infinitely large sequence, and in this case we shall use the designation yn n(n~ 1) an~2b2,

(1)

which is valid for positive a and b and for any natural number n. The examples and problems considered in this section lead to the following results: log|Q| na < a " < n\ for a > 0, | a | > l . (2) (The relation log|a| n 0, hut in Exercises 24 and 25 you have to prove it only for a > 1.) II. Control Questions and Assignments 1. Formulate the theorem on three sequences. 2. Write the first four terms of the binomial formula for the expansion of (1 + a)n. 3^_Given infinitely large sequences {«!}, {log10«}, {Vn}, {4n}, (n5}, with the aid of relation (2) find for every two sequences which of them has a higher order of growth. III. Worked Problems 1. Prove that lim —fj- = 0 for a > 0 , | a | > l . n -» o o a A It follows from the definition of a limit of a sequence that if lim |u n| = 0, then lim un = 0 as welln-> 00 n-> 00 It is therefore sufficient to prove that lim «;„=(), where n— ► 00 a wn = 1^ - . We represent wn as wn

[___2__ \ (|a | 1^a)n

and consider a sequence z„ =

) .

W e

shall prove

that lim z(l = 0. Then it will follow from the result 11-+OQ 42

of Example 5 in

2.1 that limu;n = 0. Since a n-+oo

;>0

’

| a | > 1» it follows that 3|3 > 0 such that | a |1/“ __ 1 -J- p. Applying inequality (1) to the binomial (1 -)- p)'^ we get (|a |1/a)K= (1 + P)U> n (n~ 1}- P2 V » > 2. It follows that n a | nta


2. The se­ quences {a;n}, {zn}, {f/„} satisfy the conditions of the theorem on three sequences since xn ^ z„ ^ ?/„ and lim x n = 0, 11-+OQ liin y n = 0. Consequently, lim zn = 0, and this is a -v o o

?i-►00

what wo wished to prove. We can thus write na

t),

| a | > 1- A

2. Prove that lim -^r-= 0 for | a | > l . n -> o o n l lal" and prove A Let us consider a sequence z„ = -^j— that lim zn = 0. Assume that k is a natural number such that k > | a | and that n > 2k. We represent zn as a product of ?i factors |a|n |a| |a| |a| |a| Jap n! — 1 2 2k 2A+1 ‘ ’ n * Since /c^s|a|, the fraction

and all fractions follow

ing it are not larger than 1/2. Therefore we obtain an estimate 7 1 ( 4 -)" -“ + ,= ( 2 W ) ^ ( 4 - ) n. Zn = M n! Since z „ > 0 and lim (2 lal)2*"1(4 -)” = 0, according to 71-VOO \ £t the theorem on three sequences we have lim zn = 0. an

From this it follows that lim — = 0, n->oo n
n(/t~ 1) From this we find that P „ ^ J / Vrc^2. Accord­ ing to the theorem on three sequences, it follows from the inequalities

0

\ ■ 25. Prove that lim log° n ^ 0 for < z> l, a > I . n -fo o

»

26. Prove that the following specified sequences are ii 3n infinitesimal: (a) > OO •rn£='jjip ’ Xn = Y 5 — 1. 27. Using the result

44

of Example 2, prove that

2.5. Monotonic Sequences I. Fundamental Concepts and Theorems The sequence {a:,,} is nonincreasing (nondecreasing) if •'l'n+1 ^ &n (*^n+l ^ ^n) . Nonincreasing and nondecreasing sequences are known as monotonic sequences. The sequence is increasing (decreasing) if xn+1 >. •Tn (J’n+J x n) V/J. Increasing and decreasing sequences are also called strictly monotonic sequences. Note that a monotonic sequence is always hounded at least from one side, a nonincreasing sequence is bounded from above and a nondecreasing sequence is bounded from below, by its first term. If a monotonic sequence is also hounded from the other side, then it converges, i.e. the following theorem holds true. Theorem 9. A monotonic bounded sequence converges. IT. Control Questions and Assignments 1. Formulate (a) the definition of a monotonic sequence, (b) the test for convergence of a monotonic sequence. 2. Is the boundedness of a sequence the necessary and sufficient condition of convergence of (a) a monotonic sequence, (b) an arbitrary sequence? III. Worked Problems Find the limit of the sequence {xn} which is defined by the recurrent relation: ^n+i = *n (2 — x n) (1) where is an arbitrary number satisfying the inequalities 0 < .r, < 1. A We shall first prove that the sequence {*„} is hound­ ed, namely, we shall prove by induction that V/z there hold inequalities 0< < 1. (2) For Xj inequalities (2) are satisfied by the hypothesis. yVssume that inequalities (2) are valid for the number n. \Ve shall prove that then they are also valid for the num­ ber n + 1. We write formula (1) in the form = 1 — (1 — x n)2. 45

It follows from inequalities (2) that 0 •< (i — .r,,)” < i, and therefore 0 < x n+1 1. Dividing relation (1) by x n, we get Xn+\hCn — 2 x n I. It follows that x n+1 >» xn Vn. Thus the sequence {.r„} is monotonic and bounded. Consequently, according to Theorem 9, there is a limit lim x n which will be de77—*-CO signated as a. To find a, we pass to the limit in the recur­ rent formula (1). We get lim x n+1= lima:n•lim (2—xn), or a = a (2 —a). TI-+OO

71—> 0 0

72—►OO

Hence a = 0 or a = 1. Since X] >. 0 and the sequence {xn} is increasing, we have a = 1. A IV. Problems and Exercises for Independent Work 28. Prove the convergence of the sequence {.rn}, where h= i

29. Prove the convergence and calculate the limit of the sequence {xn} if Xl = ]/a , ,r2= V a + Y Z , . . . , .rn =

/ a +V + V"a (the total number of roots is n), . . . , where a > 0. 30. Prove the convergence and calculate the limit of the sequence {xn} if it is defined by the recurrent relation (a) x n = (zn_i + xn_2)/2 Vn > 3, Xl = a, Xa = b, a b, b) zn+i = y ( xn + V n> 1, a > 0, where x1 is an arbitrary positive number, (c) xnJrl = — [xn + — ) Vra> 1, a >

,

0

where xx is an arbitrary negative number. 31. Prove that an unbounded monotonic sequence is infinitely large. 32. Prove the existence of the limit lim ( l -1- — 71— .m V ^/ 46

2.6. Limit Points I. Fundamental Concepts and Theorems Let {xn} be a number sequence. We shall consider an arbitrary increasing sequence of positive integers /f1? k 2, . . ., k n, . . . . Note that k n ^ n. From {.rn} we choose terms with the numbers /clf k 2, . . ., A’,,, . . . : • • •»

••• •

The number sequence obtained is a subsequence of the sequence {#„}. Theorem 10. I f lim xn = a, then any sequence {£fcn} 71—>00

converges to a as n oo. Theorem 11 (Bolzano-Weierstrass theorem). Any bound­ ed sequence always contains a convergent subsequence. Definition 1. The number a is a lim it point of the sequence {xn} if the sequence {a;n} contains a subsequence {^*n} con­ verging to a. We can give another, equivalent, definition of a limit point. Definition 2. The number a is a lim it point of the sequence {xn} if in any e-neighbourhood of the point a there are in­ finitely many terms of the sequence {a;n}. Remark 1. It follows from Theorem 10 that a convergent sequence has only one limit point which coincides with its limit. Remark 2. It follows from Theorem 11 that every bound­ ed sequence has at least one limit point. The largest (smallest) limit point of the sequence {:rn} bounded from above (from below), is the limit superior (limit inferior) of the sequence and is designated as lim x n (lim a:n). It is evident that if {:rn} is convergent, then lim.rn = 71 — ►OO lim xn = lim xn. If the sequence {.rn} is not bounded n-+°°

____

from above (from below), then we set lim.rn = + o o 71 (lim xn —oo). -*■ OO

71-*- «x»

47

It. Control Questions and Assignments 1. Formulaic the definition of (a) a sequence, (b) a limit point (give two definitions and prove their equivalence), (c) the upper (lower) limit ol’ a sequence. 2. Give a geometric interpretation of the definition of a limit point. 3. Is the limit of a sequence its limiting point? Sub­ stantiate the answer. 4. Given the sequences {n (—I )’1+ 1}, {/?.}, {(—l)n 1), indicate the sequence which (a) has a limit point, (b) does not have a limit point, (c) lias two limit points, (d) has only one limit point. Are there convergent se­ quences among them? 5. Prove that a convergent sequence has only one limit, point and that point coincides with its limit. Find whether the converse statement is true: “If a sequence has a single limit point, then it is convergent”. 6. Given a sequence {xn} and it is known that any neighbourhood of the point a contains infinitely many terms of the sequence and none of the intervals to which the point a does not belong, does not contain infinitely many terms of the sequence. Does it follow that lim x n = «? 71-+00

Ije*i Ijj*1 -rn = 4. Can the sequence {.r„} be (a) con­ vergent (if it can, then what is its limit?), (b) divergent? 8. Formulate the Bolzano-Weierstrass theorem. 9. bind whether the following statement is true: “If a sequence is unbounded, then a convergent sequence can be isolated from it”. III. Worked Problems 1. Prove the divergence of the sequence x„ = (—I)71. A Consider two subsequences of this sequence: x2h = 1 and x2fl^ — —1 (/.’ = 1, 2, . . .). It is evident that lim x2lt = 1, lim x2il_1 = —1. k -+ o o

h -+

OO

Thus the sequence {(—1)"} has two limit points, 1 and —1, and therefore cannot be convergent since a convergent sequence has only one limit point. A 2. Find (a) all limit points of the sequence {sin »°}, (b) the upper and the lower limit of this sequence. 48

A (a) Each of the numbers 0, ±sin 1°, ± sin 2°, . . ± sin 86°, ±1 appears in the sequence infinitely many limes since sin n° — sin (360°/) -|- »°) V«, p 6 N. There­ fore each of the indicated numbers is a limit point of the sequence {sin ?i°}. The sequence has no other limit points since if the number a does not coincide with any one of these 181 numbers, then there is a neighbourhood of the point a which does not contain any terms of the sequence. (b) From the 181 limit points indicated in (a) the point —1 is the least and the point 1 is the greatest, i.e. lim sin /z° ^ 1, lim sin n° = —1. ▲ ?}-*OO 71— >00 3. Find (a) all limit points of the sequence 0, 1, 1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5, 3/5, 4/5, 1/ 6, . . ., (b) the upper and the lower limit of this sequence. A (a) The sequence {a:,,} is the set of all rational num­ bers belonging to the interval [0, 1] which appear in the indicaled order. Since any e-neighbourhood of any real number of the interval [0, 1] contains infinitely many rational numbers (see Exercise 5 in 1.1), every point of this interval is a limit point of the given sequence. Now if the point a g [0, 1], then it is not a limit point of the given sequence since there is a neighbourhood of such a point, which does not contain any term of the sequence. (b) It is evident that lim xn = 0, lim x n = 1. A n"^ 4. Prove that the infinitely large sequence {xn} does not have a limit point. A We use an indirect proof. Assume that a point a is a limit point of the sequence {z,,}. Then there is a sub­ sequence {x/{ } such that lim x* = a. On the other 7* hand, {xh } has no limit since it is infinitely large. Indeed, since {#„} is infinitely large, it follows that VA > 0 3A: Vn > A | x n | > A_ Since k n > n and k n+i > /‘:n> it follows that VA-„ > N: \ x h n \ > A , i.e. {xh } is infinitely large. The contradiction obtained proves that {#„} does not have a limit point. A TL-+OQ

IV. Problems and Exercises for Independent Work 33. Prove that (a) an infinitely large subsequence can be isolated from any unbounded sequence, 4—01682

49

(i>) any subsequence of an infinitely large sequence is infinitely large, (c) a nionolonic unbounded sequence does not have a limit point,

(d) every bounded sequence lias an upper and a lower limit. 34. Given that the sequences {a;n} and {yn} have a li­ mit point each, give examples showing that the sequences {xn + z/n} and {xn 'yn} may (a) have no limit points, (b) have one limit point each, (c) have two limit points each. 35. Find all limit points of the sequence {;rn}, also find lima^ and lima;n if (a) *n = ( — I )71' 1(2 + ^ -) , (b) a:n = l + -— r cos^i , (c) zn = l + z lx

2

( - l )n+1+ ( - l ) n(n- |)/2,

n —1

2nn

(e) xn = 1 + n sin — , (f) *» (g)

( ~ l ) n

n

2

( l + - ) n- ( - 1 ) " + s i n ^ ,

(h) ^ = - —

s in ^ ,

(j) Xn = COSn- ~ ,

(i) Xn= V i + 2n^ i)n,

(k) xn = ( — l)n 72.

36. Test the sequence ZL n

for convergence.

2.7. Fundamental Sequences. Cauchy Condition for Convergence of a Sequence I. Fundamental Concepts and Theorems Definition 1. The sequence {.r„} is fundam ental if V e > 0 3TV such that Vn ;> N and V natural number p the inequality I x n — x n +I, \ < e holds true. This definition is equivalent to the following one. Definition 2. The sequence {a:,,} is fundam ental if 50

Ve > 0 3# such that and Vm > the inequality | .r7, — .rm | 0 3N such] that the distance between any two terms of the sequence with numbers larger than N is smaller than e. Theorem 12 (Cauchy condition for convergence of a se­ quence). For a sequence to be convergent, it is necessary and sufficient for it to be fundamental. II. Control Questions and Assignments 1. Formulate the definition of (a) a fundamental sequence (give two definitions and prove their equivalence), (b) a nonfundamental sequence (using the rule of con­ structing negations). Give a geometric interpretation of these definitions. 2. Formulate Cauchy condition for convergence of a sequence. III. Worked Problems 1. Using Cauchy condition, prove the convergence of sin k the sequence {.rn}, where xn —2 k~ / 0 and set N = [1/el. Then \fn ;> N the inequality ti M/el -| 1 >• 1/e holds inn*, whence 1 / « < k . Consequently. using inequalily (1), Vn > yV and V natural we yet | .»„ — |< 1In < e. This proves that the sequence is fundamental. ▲ 2. Using Cauchy condition, prove the divergence of the sequence {a:n}, where xn =

2

fc=i Y k A By virtue of Cauchy condition, it is sufficient to prove that the sequence {a:,,} is not fundamental. For that purpose we evaluate | x n — x n+p \. We have n+p

l^-n x n + j>I

2 /i=n+l V k ^ Yn + p In particular, for p = n we get

Vn, />£N.

K —.r2nl> -

— V». (2) Y 2 "" Y i We take e — l / ) / 2. Then VyV 3 n > yV and a natural /> such that | x„ — ,r„+|l | ^ e. Indeed, by virtue of ine­ quality (2), it is sufficient to lake any n > yV and p — ». This proves that the sequence {x„} is not fundamental. A _

IV. Problems and Exercises for Independent Work 37. Using Cauchy’s condition, prove the convergence of the sequence {#„} if (a) * . - 2 4 . (b) /l=l h= 1 n (d) xn = 2 ahY , where \q\
0. 38. Using Cauchy condition,

2

.

h=l

and |a,t| ^ M V/c,

prove that if the se-

71

quence xn = 2 ah Is convergent, then lim an = 0. A=1

71— >oo

39. Using Cauchy condition, prove the divergence of the sequence {;rn} if n

(a) *n = 52

2

a= i

n

( - 1)*,

(b) ^n =

2

/i=i

-F-

Chapter

3

Limit of a Function. Continuity of a Function 3.1. Lim it of a Function. Limit Theorems. Infinitely Large Functions I. Fundamental Concepts and Theorems 1. Limit of a function at a point. Assume that x is a numerical variable and X is its range. If we put a num­ ber y into correspondence with every number x £ X then we say that a function is defined on the set X and write y — j (x). The variable x is an independent variable (or the argument of the function), the set X is the domain of definition (or simply domain) of the function f (x), and tlie number y which corresponds to this argument x is a particular value of the function at the point x. The set Y of all particular values of the function is the set of values (range) of the function f (x). The point a (a ( X or u (f X) is a limit point of the set X if in any neighbourhood of the point a there are points of the set X different from a. In the definitions given in this section we assume that a is a limit point of the set X which is the domain of de­ finition of the function / (z). Definition 1 (Cauchy’s). The number b is the limit of the function f (.r) at the point a (as x a) if Ve >» 0 38 > 0 such that Va\ satisfying the conditions x 6 X, 0 < | x — a | < 8, the inequality | / (a;) — b | < e holds true. Definition 2 (Heine’s). The number b is the limit of the function f (x) at the point a if for any sequence {xn} con­ verging to a and such that x n £ X, x„ a, the corresponding sequence of values of the function {/ (x„)} converges to b. The notation is lim / (x) — b, or / (x) —»- b as x —*-a. x -> a It should be emphasized that the concept of the limit of a function of a point a is introduced only for the limit points a of the domain of the function. Note that in that case the function may not be defined at the point a, i.e., in general, a £ X. 53

Here are the formulations of the negations of Defini­ tions 1 and 2. Negation of Definition 1. The number b is not a limit of the function / (x) at the point a [8 lim / (x)] if x-+a

e > 0 such that V8 > 0 3x £ X for which 0 - a (xn < a), the corresponding sequence {/ (.r„)} is infinitely large. Definitions 1 and 2 are equivalent. Let the function / (x) be defined on the ray (c, ■oo). Definition 3. The function f (x) is infinitely large as x +oo if VM > 0 3A (A > c) such that Vx > A: | / (x)' | > M. Definition 4. The function f (x) is infinitely large as * +oo if for any infinitely large sequence {.+ } (,r„ > c) ^ corresponding sequence {f (xn)} is infinitely large. The notation is lim f (x) — oo. _

-V --I CO

Definitions 3 and 4 arc

equivalent.

By analogy, ive can introduce the concept of an in­ finitely large funclionLas x ->---- oo: lim / (r) = oo. If the function / (.r) is infinitely large asV % -+ o o and as x °°> then we write lim / (x) = oo. For instance, X->- o o lim x — oo. X-*OC

II. Control Questions and Assignments 1. Formulate two definitions of the limit of a function at a point. What does the equivalence of these definitions mean? 2. Using Heine’s definition of the limit of a function, prove the uniqueness of the limit of a function at a point. 3. Prove that Vj 0 lim x = x0 using (a) Cauchy’s, .V-KTo

(b) Heine’s, definition of the limit of a function. 56

4. Given a function f (x) find whether the function / (.r) is defined at the point x — 0. Is the point x = 0 a limit point of the domain of the function? Is there a limit lim / (a:)? .v->0 5. Formulate the negations of two definitions of the limit of a function at a point. 0. Formulate Theorems 2 and 3 on the limits of func­ tions. 7. Formulate two definitions of one-sided limits of a function and the negations of these definitions. 8. Are there /(3 + 0) and / ( 3 - 0 ) if /(* )= ? Is there lim / (.c)? x— >-3 9. Under what conditions does the existence of onesided limits (a limit of a function) imply the existence of a limit of a function (one-sided limits)? 10. Formulate two definitions of a limit of a function as x -v -)-oo and the negations of these definitions. 11. Prove that the function j (x) = x has no limit as x —>- —(- o o . 12.

Provo

that

lim

.r-= -|-o o .

-{-OO

13. Korin ii Iale the Cauchy and Heine definitions cor­ responding to the following symbolic notations: (a)

lim /(.I-)--oo,

(b )/(tt|-0 )

— oo,

(c)

l i m / (x) _ b,

.V-NI x~*-0° (d) lim / (x) — oo, (e) lim j (x) — -|-oo. .X— ► oo .v-v-oo 14. P r o v e t h a t Hie f u n c t i o n / (. 0 and set 5 = e. Then, if | x | < 6, then | sin x | | x | < 6 = e. And this means (ac^ cording to Cauchy’s definition of a limit of a function) that lim sin x = 0. ^ 2 . Calculate lim /(a), where / (x) = x< i * . 1 X * A Since lim (x2— 1) —0 and lim(x —1) —0, it follows 1 x-y1 57

that this limit is an indeterminate form of type 0/0 and we cannot use Theorem 2 on the limit of the quotient of two functions. We use the fact that when the limit of the function at the point x = 1 is considered, its argument does not assume a value 1. Therefore, lim / (x) — •x-vl r 2 __ \ lim (x -| 1) since / (x) = ——j—=- x -|- 1 if x =/-= 1. Lei

jc-M

x

1

{a;,,} be an arbitrary sequence converging to 1 (xn 1) and then lim (x n -f 1) — 2. This means (according to 72-vOO Heine’s definition of the limit of a function) that lim (x + 1) = lim / (x) — 2. A X-*-

1

X

-*-1

3. Calculate the limit lim 1 3 + ^ —2 A As in Example 2, this limit is an indeterminate form of type 0/0. However, as distinct from Example 2, here we cannot directly “cancel” x — 1 out of the numer­ ator and the denominator. Therefore we first transform the function by multiplying the numerator and denomi­ nator by 0^3 + x + 2), i.e. by an expression conjugate to the numerator. We obtain / 3 + J-2

x—l_______

x~ i (x—i) ( 1^3 + x-{-2) Since, when this limit is considered, the argument x does not assume the value x = 1, wo get the following expression when cancelling out x _1: r x —2 lim 1^3x “— lim — 1 __ 1 X- I x-» i L3-I-.T-I-2 4. Prove that the Dirichlet function 0 if x is an irrational number, D(.r) = 1 if x is a rational number does not have a limit at any point. A We shall prove that at an arbitrary point a the function D (x) does not satisfy Heine’s definition of a limit of a function. To do that, we indicate two sequences {.r„} and {x'n j which converge to a and are such that, lim D (x„) =£ lim D We shall first consider the 71—►OO

7l-> 00

sequence {x’n} of palional points, which converges to a. 58

For this sequence D (xn) = 1 Vrc and therefore lim D (x n) = 1. Then we consider the sequence {x'n} —►o o of irrational points, which converges to a. For this se­ quence D (x'n) = 0 Vn and therefore lim D (x'n) — 0. ll

71 -►OO

Thus lim D (x n) ^ lim D (x'n). Hence it follows that 71 -*-CV>

71->00

the limit of the function D (x) at the point a does not exist. A 5. Let us consider the set of all irrational numbers belonging to the interval (—1, 1). We designate it as I ir. We define the function / (x) on the set I iT: / (x) = 1 if x 6 Prove that lim / (a) = 1, where a is an arbitrary x -* a

point from the interval [—1, 1] (rational or irrational). A Let a £ 1—1, 11- The point a is a limit point of the set I ir. We use Heine’s definition of a limit of afunction. Assume that {xn} is an arbitrary sequence of points of the set 7,r, which converges to the point a (xn a). By the hypothesis / (x n) = 1 Vx„ 6 Ar. and therefore lim / (xn) = 1, and, consequently, lim / (x) = 1. A oo x-+ a 6. Prove that the function sin x does not have a limit as x + oo‘ A We shall prove that this function does not satisfy Heine’s definition of the limit of a function as x — +oo. To do that, we indicate an infinitely large sequence {.ra} such that the sequence {sin .rn} diverges. We set x n = ^(2n + 1). Then lim x n = +oo and the sequence T t~ y

{sin x n} = —1, 1, —1, 1, . . . diverges. It follows that the function sin x does not have a limit as x —>►+°°- A 7. Let for £ < 0, / (*) = { X. v 1 { sm x for x > 0 0). Is there a limit lim / (.r)? (/ (x) is not defined for x x-*-0 A We calculate the one-sided limits of the function / (x) at the point x = 0 using Theorem 5 for the functions y — sin x and y = x at the point a = 0: f(a + 0)= lim sin x = lim sin x = 0, ic-»0-f0 oc-*0 {(a —0) — lim x = lim x —0. *-►11-0 *->-o

According to Theorem 4, it follows that there is a lim it lim / (x) and it is equal to zero. A .•c->0 100x2 + 1 8. Calculate lim f(x), where / (x) x 2 + 100 • X -> o o

A This limit is an indeterminate form oo/oo since the numerator and the denominator arc infinitely lai’ge func­ tions as x oo. We represent / (x) in the form 100 + 1/X2

/(*) l + ion/x3 (x =/=0). Since lim (\/x2) — 0, we can use Theorem 2 (as x->- oo) and obtain lim (100+ l/x2) X -* -o o

/ (*) =

~~T = 1 0°- A

lim (l + l o o / x 2)

X-VOO

Examples 2 and 8 allow us to formulate general rules for calculating limits of the form lim R (x) and X— ► (i

lim R (,x). Here R (x) is a rational function (rational v fraction), i.e. R (a.) = P n (x)/Qm (*■), where P n (a) and Qm (x) polynomials of degrees n and m respectively.

X-+oa

If lim Qm (*) ^ Q m (a) =£ 0, then lim R (x) - - 4 ^ . Q m (a )

If lim Qm (x) = 0 and lim 1 \ (x) =jt=0, then X-* ;n, lim R (x) = < a0/ 60 for n = m, „0 for re 0 Vrc. Since lim (—l/x n) = — o o , we have n->°° lim e /xn = 0 . Consequently, lim e~1/x = 0 Let n_*co .. . x-0 +0 us consider now an arbitrary infinitesimal sequence {x'n} such that x'n < 0 Vn. Since lim (—llx'n) = +oo, n 71—^00 it follows that lim e~1/xn= -f oo. Therefore lim 71-+0O x_yn_ n 4-co, i.e. / (0 — 0) — +oo. Thus the limit of / (x) on the left of the point x = 0 does not exist and, hence, x = 0 is a point of disconti­ nuity of the second kind. (c) We shall prove the continuity of / (x) at the point a =£ 1. We lake e < | a — 1 |, e > 0 . Then the neigh­ bourhood of the point a does not include the point x = f 60

if e < | a — i |. In this e-neighbourhood / (.r) coincides either with the function cp (xj = x if a < 1 or with the function i|) (x) = In x if a > 1. Since these basic ele­ mentary functions are continuous at the point a, the fund ion / (;r) is continuous at any point a 1. Wc test the function / (.r) for continuity at the point a = 1. For that purpose, we calculate its one-sidecl limits at this point utilizing the continuity of the functions cp (x) and \|> (:r) at the point 0, a=fci. x-*- 0

z

A (a) We represent the function in the form In (1 + x)l/x— In y, where y = (l + a-)1/*. Since lim (1 +a:)‘/*= e and the function In y is continuous at x-*-0 the point y —e, it follows that lirn In (1 + :r)1/'x= In o= 1. x-0

(b) Let us consider the function y = cp (a:) = ax — 1. It is continuous at the point x = 0 and y (0) = 0. In this case * = !oga C1+ !/) and We calculate lim ■ in item (a): lim

a*x

^ ^

1

y logo (l + i/) *

using the result obtained

y____________i/lna

V-0 lo8 a ( l + J/)

In (1+J/)

= In n ----- —i------- -- Ip a li,n In o

y

1

„

“ In a •

67

We shall-consider now the function / (//) continuous at the point f/ = 0: ;/ for ;/ (), lo g „ (1 //) /('/) for y —0. In a According to Theorem 8, the composite fund ion r

/(* r ( •'))-

ax — 1

x In a

for x =j£=(I, for x - 0

is continuous at the point x —- 0. Therefore, lim ax —1 I) In a. A IV. Problems and Exercises for Independent Work 15. Test the function / (x) for continuity and indicate the kind of its points of discontinuity (see Exercises 1-4): (a) / (x) = x sin (l/.r), (b) / (x) = sin (1/x), (c) / (x) -{•*'}> x2 if x is an irrational number, w /( * ) = { f jif x is a rational number, x2-\-2 (e) /(x) (f) j (x) —arc tan (1/x), X'1— f*1

Hi. . , (a)

-*> ’ (h) / (x) = In (x -|-l) x2 for 0 ^ . r < l , /(x) = 2 — x for I < j ^ 2, x for | x | < l , /(* )1 for | x | > 1, cos (nx/2) for | x | < l , f(x) |x — 11 for |x| > 1. Prove th at cx—1 . n\ (1-\-x)a —1 h m -------1, (b) lim - ----= a,

w

x- - 0-

(g) /(*) = (i) (j) (k)

1_ ex/0 % 1—cosli x 1 (e) lim ----- -5------2 x-+i) x 08

(x— 3)

3.3. Comparison of Infinitesimal Functions The Order Symbol “o” and its Properties I. Fundamental Concepts and Theorems

1. Comparison of infinitesimal functions. The func­ tion a (x) is infinitesimal as x a (at the point a) if lim a (a) = 0. Let a (.r) and p (a:) be two infinitesimal X-VfZ functions as x a. The functions a (x) and p (x) are (a) infinitesimal junctions of the same order as x a (at the point a) if lim 0, PW (b) equivalent infinitesimal functions as x — a (at the point a) if lim

x-+a P(*)

~

1

(the notation is a ~ P as x-»-a).

If lim -fTT^r- = 0, then we say that a (#) is an infinix-+a P W tesimal of a higher order of smallness as x a (at the point a) than [3 (x) and write a = o ((3) as x ~+a (a is equal to “o small” of (3 as x ->a). For example, x2 == o (x) as x ->0. Similar definitions are true for the cases x —^a -)- 0, x a — 0 and x oo. Il should be borne in mind that the equalities which contain “o” are conventional. For instance, the equality x2 o (x) holds true as x -+0 but the equality o (x) == x2 does not since the symbol o (a:) denotes not a specific function but any function which, as x -> 0, is an infini­ tesimal of a higher order of smallness than x. There are infinitely many functions of this kind. In particular, any function xv (where p > I) is o (.r) as x -^0. Thus the equality x2 = o (x), as x -> 0, means that the function x1 belongs to the set of infinitesimal functions of a higher order of smallness, as x —v0, than x.Therefore, the reverse of this equality, i.c. o (x) = x2, does not hold Irue since the whole set of functions o (x) dees not reduce to a single function x2. 2. The properties of (lie symbol “o” Theorem 10. Let ax (x) and a2 (x) be two arbitrary in­ finitesimal functions as x ->- a such that (x) = o ((3) 69

and a 2 (x) = o (p). Then a, (x) -|- a 2 (x) = o ((1) ~oo. Give examples of functions of this kind. 2. formulate the definition and give examples of an infinitesimal function a U) (a) of the same order as the function [J (x) at the point a, (b) equivalent to the function p (x) at the point a, (c) of a higher order of smallness, as x than P (*)• Interpret 1lie symbolic notation r/. — o (P) as X -+Ct. 3. Give examples ol Iunctions a (x) for which the fol­ lowing equalities hold true: (a) a (x) = o (x) as x - > 0; 70

(b) a (z) = o (Vl — x) as x -^-1 — 0, (c) a (x) = o (i/x2) as x —»- oo. 4. Provo iliaL z 3 = o (z2) as x ->-0. Find whether the equality z 3 = o (p) holds true as j -^-0 if (a) P (x) = x, (b) p (z) «= x2 Y \ x |, (c) p (x) = x3 Y \ x |, (d) P (z) = .r2sin x. 5. Prove that (x — l )2 = o (x — 1) as x ->1. Find whether the equality (x — I )2 = o (p) holds true as .r ~>1 if (a) p (x) = (x - l)3, (b) p (x) = sin (x - I)2* )2 (c) p (x) = (*—I In x 6. Prove that I /#4= o (I/#3) as # — oo. Does the equality l/rr4= o(P) hold true as x oo if (a) P W = - 3 r (*=■!. 2). (b) P M = - J r -

(e) P I3-)

(,T —

arctan (1/x)

7. Are the functions sin z and x equivalent infinitesi­ mals as x —vO? Prove that sin x — x = o (x) as x —>-0. 8. Using the properties of the symbol “o”, write an equality of the form a (x) = o (xh) as x ->-0 for the func­ tion a (x) if a (x) = o (x2) + o (z2), a (x) -o (x ) —o (z), a (x) —5o (x), a (z) = o (3z2), a (z) = (o (x))3, a(x) = xo(x), a (x) = , a (x) = o ( —x + 2x2+ z4), a (z) = o (o (z2)), a (z) —o (z + o (x)). 9. Using the properties of the symbol “o”, write an equality of the form a (x) = o (i/xh) as x oo for the . function a (z) if a (z) = 0 (1/z) o (1/z), a (z) = 1000o (1/z), a (z) = o (1000/z), a (z) = (0 (i/YJx\))2, a (z) = z2o (1/z3),

a

« n

x-+ a

indeterminate form of type 1°°. The second remarkable limit is an example of an indeterminate form of this type: lim

(1

+

= e.

x-*0

If we represent uv as e® ln u, then each of the indeter­ minate forms (0°, oo°, 1°°) reduces to an indeterminate form of type 0 *oo for the function v In u. If, in this case, lim i;lnu = ft, then limzi" —cb. x-> a

78

x-> a

. Control Questions and Assignments 1. Write asymptotic formulas for the functions sin x, tan x, cos./-, In (1 + .r), ev, ax, (1 -|- .r)n, sinh x, lanli x, cosli x as x —vO. 2. Write asymptotic formulas with the remainder of the form o (:c“) as x —vO or o (i/xa) as x — oo (a > 0) for the composite functions sin p, tan p, cos y, In (1 + p), gv, ay, (\ -| p)“, sinh p, lanh p, cosh y if (a) p = 3.r and x —vU, (h) p — ]/.r and x -| 0, (c) p — x3 and x —y 0, (d) p — l/.r and x — oo. 3. Write asymptotic formulas willi the remainder of the form o (1/«“) (a > 0) for the sequences sin x„, tan x n, cos x„, In (1 j- a,,), e*", a*", (1 -|- xn)a, sinh x n, lanh x„, cosh x n if (a) x n — Mn, (b) x n = 1In2, (c) x n = 11

In (l + ± ) , (d) xB = eVB - I4. Give a derinition of an infinitesimal sequence {an} of a higher order than {1hi) as n ->~oo. Give a definition of the infinitely small function a (:r) of a higher order than i/x as x —>- oo. Write the symbolic notations for a n and a (x). 5. Of what order of smallness are the sequence a n = no (1hi)2 as n —v oo and the function a (x) = xo (i/x2) as x oo as compared to {1/w} and (i (a:) = i/x respec­ tively? 6. Using the asymptotic formula IV and Heine’s definition of the limit of a function, prove that e1/,l = 1 + ^ + ° ( ¥ ) as n ~ ^ Q°7. Find whether the following equalities hold true: y i + %2 c08h T

i + y x2+ o (x2) as + T - S - |- ° ( p - ) as " ^ ° ° -

= 1

Substantiate the answer. 8. Enumerate possible cases for calculating the limits of exponential power functions. Give examples of three types of indeterminate forms for those functions. 9. Calculate the following limits:

0 ° £ +„(,*> - - | H j 2 2‘ m cr& T +: iimr 1

x-+ x-»00

— _ + 11in — 2 80

*-h> *'

fl

*

Here we have used the fact that according to the defini­ tion of the symbol “o" lim = 0. A x-*0

x~

2. Calculate lim a ~~a“ (a > 0 ). x-»« x ~ a A We set y - - x — a. and then we can write the limit as aM*a —aa lim --------— •aa

y~>o

Since ay

y

01.

lim

»/->o

a.v — 1 y

.

l + i / In a + o (y), we have a" lim ———= aa lim y-* 0

y

V~* 0

= aa ( In a + lim '

y-+o

y In a+o( y)

y j = aa In a.

V

Thus, lim ——— = aaIn a. A *-« x~ a ____ 3. Calculate lim sin (n l/ « 2+ 1 ). 71—> ° ° A Using the asymptotic formula V for x = lI n 2, and a = 1/2, we obtain V nZ+ 'l = n ( i + - ^ ) l,2 = n ( l + Y - - ^ + 0 { w ) ) = n+ i + ° ( i ) Hence sin (n ]/n 2+ 1) = sin (jira + £ r + ° ( 4 ) ) = ( . - 1)” s i n ( ^ - + o ( l ) ) . The sequence {( —1)"} is bounded and {sin (^-- + • ( 4 ) ) } is infinitesimal, and therefore the product of these two sequences is an infinitely small sequence. Thus lim sin(n]/rn2+ l) = 0. A «- 4. Calculate lim (cos x) x. x-t-0

A This limit is an indeterminate form of type 1°° since lim cos a := l, lim cot2x = oo. We write (cosx)Cot** x-o x-o ' 6 -0 1 6 3 2

81

in the form ecot2*•incos x and calculate L = lim cot2x •In cos x. x-*0

For that purpose, we write an asymptotic expansion for In cos a: and sin2# as # —>-0: In cos x —In ( t —

+ o (x2) j = —— + o (x2),

sin2x = (x + o (#))2=

-f- o (x2).

#2

Using these relations, we find that L = lim cos2x •lim X -v O

3f->0

+ 0(#2)

#2

2

•

Thus the required limit is equal to eL = e-1/2. A 5. Calculate lim /J_\tan(1/n^ 7 i— >oo Vn / A This limit is an indeterminate form of type 0°. To calculate it, we write ^JL^tan(1^n) jn form etan(i/n).in(i/n) and calculate L =■ lim ( tan —x In —] . n~>oo ' n nJ We have

= — lim — lim [in n-o ( —) 1 n-*oo n 7i— voo L \ nIJ = 0— lim lim — \ = n-VOO n 71-VOO 1 n

0

since lim -^ -^ = 0 (see 2.4) and lim = 0. n-t-co ^ n->oo t/n Thus L = 0, i.e. the required limit is equal to 1. A IV. Problems and Exercises for Independent Work 23. Write the asymptotic expansion for the following functions, as i - ^ 0, with the remainder of the form o (xa), where a ^ 0: (a) sin2(5 Y~x), sin2(5 Y~x+ x) ( x > 0), (b) cos (4x2), cos (4x2+ x), 82

(c) e2\ e2a:+

( x > 0), (d) ln ( l—x^), in[(l —xl + x),

(e) 3 - 3/2 7 - ^ , 3 - 3/ 2 7 - i f ] / i (x > 0 ), (f) 2X\ 2X*+X\ (h) cos ]/ sin x, In cos(2x + x2), (g) In cos 2x, co sh )/sin x (x ;> 0), _ (i) lo (£ M -j/J) 0), (j) (k) ^ c o s y x ( x > 0), (1) cosx cosx2—1. 24. Write the asymptotic expansions for the functions, as x 2, with the remainder of the form o (x — 2)“, where a > 0 : (a) sin (x —2)2, (b) (3 —x)P, (c) In (x — 1), (d) cos ( ) , (e) tan (ni2), (f) x — 1 — 1/ x — 1, (g) xx — 4. 25. Write asymptotic expansions for the functions, as x —>- oo, with the remainder of the form o (l/xa), where a ^ 0: (a) y x2+ x —x, (b) ^'x-’+ x—x, (c) V {x + 1) (f; "t" ( l “t~3) (^_+ 4) —x, (d) sin ( i / \ / x2), s in h (l/f/x y (e) cos(l/x2), (f) S1/*, (g) lncos( 2/x), lncosh( 2/x), (h) q1IV~*— 1 ( x > 0). 26. Writo asymptotic formulas for sequences with the remainder of the form o(l/na), where a ^ O : (a) n3+ n 2—n, (b) 21/n+ 7 ‘/n—2, (c) s in (l/]/« ). 27. Calculate each of the following limits: j ix

(a) lim

sin( * ~ f )

cos x — cos 3x

(b) lim - [ 3 2 x-+ n/3

COS

X

(c) lim (1 —x) tan — , (d) lim Y F ^ - V H I L x-+ 0

(e)

(m, «€N),

x

lim (\/ 1 +

x

+

x 2

— ) / l — x + x2),

r C -* --fo o

(f) lim

V

COS

X— y

COS

X

sin2 x

i-VH (h) l i m ( - ± |) 1 * as (i + x

6*

, .

(g) hm x-*-a x -» - +

In x — In a

x -a

, x - ^ 1, x-*--|-oo,

0

83

ltan 2x

(i) J i ? j tan ( f + *)]' (i) i s . (S t s )1 (k) lim (sin .t)1™*,

(1) lim »Xcosh("/»»)-i), 71— oo

(m) lim ( T ' + ' / i )" ( a > 0 , b > 0 ) . 7i—►oo

\

a

/

'7

28. Calculate each of the following limits:

.(a)

( 5C->0

(b) lim j T * + * - g ± * ^ x->0 L - l / 1- - ^ ^

„ 6N),

X

i S In cos 3* ’

^

(')

(c) sin4 3 \ f x

cxi —e3x i^ain(**/ 2)—sin* ’

) I

t

a

( « > 0),

In tan [ -j- j-ax ) (c) lim U 1 sin bx x-*-0 /i\ i- a*+/iax”^—2ax , ^ m (d) lim ---- — h-* 0

(0) l i x-0

5

-------- ( a > 0 ) ,

n

m

x

0),

In (c*2-f-2 Yx) 0

( c > 0),

tan Y x

/tA i. ’

‘Y x —\f~x

(h)i

(i) l i m [ c o s ( 2 n ( ^ ) a ) ] * J , (j) n m t a n « [ ^ A + ( l - | - | ) a] . (k) lim (sin -t- -f cos /i\ ()

i*

In (x®-j—g*) In (x4+ e « j- ’

, / \ ]• (,n) ^

In (x^-j- C'^) In (x4+ c2x) *

Chapter 4

Derivatives and Differentials 4 .1 . The D eriv a tiv e of a F u n ction . D ifferentiation R u les I . F u n d a m e n t a l C o n c e p ts a n d

T h eorem s

1 . D e f in it i o n o f t h e d e r i v a t i v e . L e t th e f u n c t io n y = / (x) b e d e fin e d in a n e ig h b o u r h o o d o f t h e p o i n t x 0. T h e increment o f t h i s f u n c t i o n a t x 0 is a f u n c t io n o f t h e a r g u ­ m e n t Ax: Ay = f (x0 -|- Ax) — / (,r„).

divided difference — - is a ls o a f u n c t io n o f t h e a r g u m e n t Ax. The

85

Definition. The derivative of the function y — f (x) at the point x0 is the limit lim {provided thatit exists). The derivative of the function y = / (x) at the point x0 is designated as /' (x0) or y' (x0). The operation of finding the derivative is differentiation. 2. Table of derivatives of basic elementary functions I. (xaY = ax'1-1 (a is any number). II. (sin x)' = cos x. III. (cos x)' = —sin x. 1 1 IV. (loga x)'=-^j— , in particular, (lnx)' = — (x > 0 ), x in cl x V. (a1) ' = a* In a, in particular, (ex)' = ex. VI. (tanx)' = -^— - (x = ^ -J + Jtra, n £ z ) . VII. (cotx)' = — VIII. (arcsin x)' = IX. (arccos x )' = —

{x=£nn, n£Z). ( — 1 < x < 1).

V 1—x 2

y —

^

( — 1 < X < 1).

X. (arctan x ) ’ — -jqrp" . XI. (arccotx)' = —— XII. (sinh x )' = cosh x . XIII. (cosh x )' = sinh x . XIV. XV. ( c o t h i ) ' =—

-0

Ax

then it is called a right-hand (left-hand) derivative of the function y = / (x) at the point x0 and is designated as f (^0 + 0) [/' (x0 — 0)1 respectively. If there are /' (x0 + 0) an^ /' (x0 — 0) and they are equal, then there is f (x0) and it is equal to /' (x0 + 0). Conversely, if there is /' (x0), then there are f (x0 + 0) 87

and f (x0— ), and in that case /' (x0 + 0) = /' (x0— 0) = r (*«)• 6. Rules of differentiation Theorem 2. If u (x) and v (x) have derivatives at the point x0, then the sum, the difference, the product and the quotient of these functions (the quotient under the condition that v (;r0) 0) also have derivatives at the point xQ , and ' the following equalities hold true at the point xQ\ (lZ+ u)' = u' + v \ (u— v)’=^ u — v', , \t / , / u ' v — uv' (uv) = UV + UV, (/ -uJ\ ' --------. 7. The derivative of the inverse function Theorem 3. If the function y = f (x) is strictly monoton­ ic and continuous in a neighbourhood of the point xQ, has a derivative at the point x0 and f (x0) 0, then there is an inverse function x = f-1 (y) which is defined in a neigh­ bourhood of the point y0 — / (a:0) and has a derivative at the point y0; in that case (I-1 (y0)Y = i / / ' (*o).

(D

The physical interpretation of formula (1) is as fol­ lows: the derivative (/-1 (y0))' is the rate of variation of the variable x relative to the variation of the variable y, and /' (.r0) is the rate of variation of the variable y rela­ tive to the variation of the variable x. It is clear that these quantities are mutually inverse. 8. The derivative of a composite function Theorem 4. I j the junction t = cp (x) has a derivative vz (*) for 0 < t < 1/ 2, vx (t) < v2 (t) for t > 1/ 2. The average velocity of the first point on the time interval 0 ^ 1 is ulay= ^ Similarly, v2av = »»(*)—MQ) = 2. Thus vt ay < v2av. A 3. Derive an equation of the tangent to the graph of the function y = cos x at the point with abscissa x — jt/6. A We have x0 = ji/6, / (x0) = cos (ji/6) = Y 3/2, /' (x0) = — sin (n/6) = — 1/ 2. Therefore the required equation of the tangent can be written in the form y

VI

1 2

2

/ \

n\ 6/ ' ^

4. Find the one-sided derivatives of the function / (x) = | x — xx | g (x) at the point x0, where g (x) is a function continuous at the point x0. Does the function / (x) pos­ sess a derivative at the point x0? A For Ax > 0 the increment of the function at the point x0 has the form Ay = / (x0 + Ax) — / (x0) = | x„ -f- Ax — x0 | g (x0 + Ax) — 0 = g (x0 + Ax) Ax, whence ^ |- = g(a:0-)- Ax). Since g(x) is continuous at the point x0, we have lim

A1 - + + 0

= g (x0). Thus / ' (x0+

) = g(*o)Similarly, for Ax < 0 we get Ay whence

0

— g (x0 + Ax) Ax,

lim - ^ -= lim (—g(x 0-f-Ax)) = —g(x0), Ax-f-0 a x

Ax-y-0

i.e. /' (x0 — 0) = — g (Xq). If g (x0) ^ 0, then /' (x0 + ) f (x0 — 0), and, hence, the function / (x) does not have a derivative at the point x0. Now if g (x0) = 0, then /' (x0 -f 0) = /' (x0 — 0) = 0 and, consequently, the 0

93

function / (x) possesses a derivative at the point x0 and /' (x0) = 0. A 5. Calculate the derivative of each of the following functions: ,

x2 sin x

v

.

^ n

. . .

(a) y = Id x (* > °» (b) y = cos (2X — x3) (—oo < i < oo). A (a) Using the rules of differentiation of the product and the quotient and the table of derivatives, we get V' (*) =

(x2 sin x)f In x — x2 sin x (In x)' In2 x 1

(x2 cos x + 2x sin x) In a:—x2 sin x — In2 x _ x (x cos x + 2 sin x) In x — x sin x ~~ In2 x

( x > 0, x ^ i ) .

(b) We can represent the function y = cos (2* — x3) in the form y = cos t, where t = 2* — x3. Using the rule of differentiation of a composite function, we get y'(x) = (c o s i)'|t=2*_xJ(2:c- ^ ) '

= —sin (2X—x=>) (2*In 2 —3x2) 6. Find the derivative y (x) of a:2sin (l/a:) for y 0 for

-I

( —o o < x < o o ). A the function x^O , x=0

and find whether y' (x) is continuous at the point x = 0. AFor x 0 the derivative y' (x) can be found by differentiating the function x2 sin (i/x) according to the rule of differentiation of a product. This yields y' (x) = 2x sin (l/x) — cos (1fx) (x =£ 0). The expression obtained is not defined for x = 0. This does not mean, however, that y' (0) does not exist since the expression for y' (x) was obtained under the condition x ^ 0. To find y' (0), we use the definition of the deriva­ tive. The increment Ay of the function y (x) at the point x = 0 is (Ax)2 sin (1/Ax) and therefore lim -jrj: = Alim Ax sin t - . li

Ajc-t-0 a x

94

Ax-*»0

A*

= 0, i.e.

m'(0)

o

\ /

= 0.

Thus y' (x) exists at all points: 2x sin (1/a:) —cos (l/a:l for x^=0, for x = 0. 0 To test y' (x) for continuity at the point x = 0 we consider lim y (x). It is clear that lira 2x sin (1/a;) x-vO x-**0 = 0 but lim cos (1/a:) does not exist. Therefore lim y' (x) X-*0 3C->0 does not exist either. Thus y' (x) is discontinuous at the point x = 0 which is a point of discontinuity of the second kind of the function y' (x). A 7. Prove that the equations x = cos t, y = sin t (O^L t ^ .n ) define parametrically the function y = f (x). Find the derivative /' (x) of this function. A The function x = cos t is strictly monotonic (decreas­ ing) on the interval and, consequently, has an inverse. Substituting this inverse function into the equation y = sin t, we get a function of the form y = / (x). In this case the inverse function is in explicit form: t = arccos x and therefore, for / (a:), we get an expression / (a:) = sin (arccos a:) (— 1 < a:< 1). We can also write this function in the form / (a;) = y 1 — x2 (—1^ a :^ 1) (explain why). Calculate the derivative /' (x) by two techniques: (a) using the explicit expression, (b) using the formula for the derivative of a function defined para­ metrically. We have (a) / ' ( * ) = - y = r

(-K * < 1 ),

(b) = 7 ts* n)Since cost = x, sinf ~ Y 1—cos2f = ]/1 —a:2for the second expression for f (x) yields the first: / ' (x) = ---- (a r^ d h l, or — 1< a: < 1). A V1

x2

. Prove that if the vector functions r± (t) and r 2 (£) have derivatives, then the following formula holds true for the derivative of the scalar product (rx (t) r 2 (£)): (ri (0 r2 (*))' = f t (0 r 2 (t)) + f t (t) r; (t)). A Let Tj (t) = iarx (t) + jyj (t) + kz1 (t), r 2 (t) = i*2 (t) + jy2 (