Problems in Calculus of One Variable - I. a. Maron_djvu

8/17/2019 Problems in Calculus of One Variable - I. a. Maron_djvu

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I.A.Maron

Problemsin

Calculusof

One Variabl

H. A. MAPOH

HHOOEPEHUHAJlbHOEW MHTErPAJlbHOEHCMHCJ1EHMEB nPMMEPAX H 3AiXAHAX


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First published 1973Revised from the 1970 Russian edition

0223-200041(01)-73

Ha aneAUucKOM R3biKe

Contents

From the Author 9

Chapter I. Introduction to Mathematical Analysis 11

§ 1.1. Real Numbers. The Absolute Value of a Real Number 11

§ 1.2. Function. Domain of Definition 15

§ 1.3. Investigation of Functions 22

§ 1.4. Inverse Functions 28

§ 1.5. Graphical Representation of Functions 30

§ 1.6. Number Sequences. Limit of a Sequence 41

§ 1.7. Evaluation of Limits of Sequences 48

§ 1.8. Testing Sequences for Convergence 50

§ 1.9. The Limit of a Function 55

§ 1.10. Calculation of Limits of Functions 60

§ 1.11. Infinitesimal and Infinite Functions. Their Defini-tion and Comparison 68

§ 1.12. Equivalent Infinitesimals. Application to Finding

Limits 71

§ 1.13. One-Sided Limits 75

§ 1.14. Continuity of a Function. Points of Discontinuity

and Their Classification 77

§ 1.15. Arithmetical Operations on Continuous Functions.

Continuity of a Composite Function 84


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§ 1.16. The Properties of a Function Continuous on a Clo-sed Interval. Continuity of an Inverse Function 87

§ 1.17. Additional Problems 91

Chapter II. Differentiation of Functions 93

§ 2.1. Definition of the Derivative 98

§ 2.2. Differentiation of Explicit Functions 100

§ 2.3. Successive Differentiation of Explicit Functions.

Leibniz Formula 107

6

Contents

§ 2.4. Differentiation of Inverse, Implicit and Parametri-

cally Represented Functions Ill

§ 2.5. Applications of the Derivative 115

§ 2.6. The Differential of a Function. Application to App-roximate Computations 122


Chapter III. Application of Differential Calculus to Investigation of

Functions 131

§ 3.1. Basic Theorems on Differentiable Functions .... 131

§ 3.2. Evaluation of Indeterminate Forms. L'Hospital's Rule 138§ 3.3. Taylor's Formula. Application to Approximate Cal-

culations 143

§ 3.4. Application of Taylor's Formula to Evaluation of

Limits 147

§ 3.5. Testing a Function for Monotonicity 148

§ 3.6. Maxima and Minima of a Function 152

§ 3.7. Finding the Greatest and the Least Values of a

Function · · 159

§ 3.8. Solving Problems in Geometry and Physics .... 162


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§ 3.9. Convexity and Concavity of a Curve. Points of

Inflection 166

§ 3.10. Asymptotes 170

§ 3.11. General Plan for Investigating Functions and Sket-ching Graphs 174

§ 3.12. Approximate Solution of Algebraic and Transcen-dental Equations 183


Chapter IV. Indefinite Integrals. Basic Methods of Integration .... 190

§ 4.1. Direct Integration and the Method of Expansion 195

§ 4.2. Integration by Substitution 199

* 4.3. Integration by Parts 202

§ 4.4. Reduction Formulas 211

Chapter V. Basic Classes of Integrable Functions 214

§ 5.1. Integration of Rational Functions 214

§ 5.2. Integration of Certain Irrational Expressions . . . 219

§ 5.3. Euler's Substitutions 222

§ 5.4. Other Methods of Integrating Irrational Expressions 224

§ 5.5. Integration of a Binomial Differential 228

§5.6. Integration of Trigonometric and Hyperbolic Functions 230

§ 5.7. Integration of Certain Irrational Functions with the

Aid of Trigonometric or Hyperbolic Substitutions .... 237

Contents

7

§ 5.8. Integration of Other Transcendental Functions . . . 240§ 5.9. Methods of Integration (List of Basic Forms of In-tegrals) 242

Chapter VI. The Definite Integral 247

§ 6.1. Statement of the Problem. The Lower and Upper


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Integral Sums 247

§ 6.2. Evaluating Definite Integrals by the Newton-Leib-niz Formula 256

§ 6.3. Estimating an Integral. The Definite Integral as a

Function of Its Limits 262

§ 6.4. Changing the Variable in a Definite Integral . . . 275§ 6.5. Simplification of Integrals Based on the Properties

of Symmetry of Integrands 288

§ 6.6. Integration by Parts. Reduction Formulas .... 294

§ 6.7. Approximating Definite Integrals 301


Chapter VII. Applications of the Definite Integral 310

§ 7.1. Computing the Limits of Sums with the Aid of De-finite Integrals 310

§ 7.2. Finding Average Values of a Function 312

§ 7.3. Computing Areas in Rectangular Coordinates ... 317§ 7.4. Computing Areas with Parametrically Represented

Boundaries 327

§ 7.5. The Area of a Curvilinear Sector in Polar Coordinates 331

§ 7.6. Computing the Volume of a Solid 336

§ 7.7. The Arc Length of a Plane Curve in Rectangular

Coordinates 345

§ 7.8. The Arc Length of a Curve Represented Parametri-cally 348

§ 7.9. The Arc Length of a Curve in Polar Coordinates 351

§ 7.10. Area of Surface of Revolution 354

§7.11. Geometrical Applications cf the Definite Integral 360§ 7.12. Computing Pressure, Work and Other Physical

Quantities by the Definite Integrals 367

§ 7.13. Computing Static Moments and Moments of Inertia.

Determining Coordinates of the Centre of Gravity .... 372


Chapter VIII. Improper Integrals 387


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§ 8.1. Improper Integrals with Infinite Limits 387

§ 8.2. Improper Integrals of Unbounded Functions .... 397§ 8.3. Geometric and Physical Applications of Improper

Integrals 409


Answers and Hints 418

In fond memoryof my parents

From the Author

This textbook on mathematical analysis is based on many years'experience of lecturing at a higher technical college. Its aim is totrain the students in active approach to mathematical exercises,as is done at a seminar.

Much attention is given to problems improving the theoreticalbackground. Therefore standard computational exercises are supple-mented by examples and problems explaining the theory, promo-ting its deeper understanding and stimulating precise mathema-tical thinking. Some counter-examples explaining the need for cer-tain conditions in the formulation of basic theorems are also in-cluded.

The book is designed along the following lines. Each sectionopens with a concise theoretical introduction containing the prin-cipal definitions, theorems and formulas. Then follows a detailedsolution of one or more typical problems. Finally, problems with-

out solution are given, which are similar to those solved butcontain certain peculiarities. Some of them are provided with hints.

Each chapter (except Chap. IV and V) closes with a separatesection of supplementary problems and questions aimed at reviewingand extending the material of the chapter. These sections shouldprove of interest to the inquiring student, and possibly also tolecturers in selecting material for class work or seminars.

The full solutions developed in the text pursue two aims: (1)to provide lecturers with a time-saver, since they can refer thestudents to the textbook for most of the standard exercises of acomputational character and concentrate mainly on the solution

of more sophisticated problems, thus gaining time for more rewar-

10

From the Author


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ding work; and (2) to meet the needs of those who are workingon their own or following correspondence courses, providing a sub-stitute for the oral explanations given to full-time students.

The student will find the book most useful if he uses it acti-vely, that is to say, if he studies the relevant theoretical materialcarefully before going on to the worked-out solutions, and finallyreinforces the newly-acquired knowledge by solving the problemsgiven for independent work. The best results will be obtainedwhen the student, having mastered the theoretical part, immedia-tely attacks the unsolved problems without referring to the textsolutions unless in difficulty.

Isaac Mar on

Chapter

INTRODUCTION

TO MATHEMATICAL ANALYSIS

§ 1.1. Real Numbers.

The Absolute Value of a Real Number

Any decimal fraction, terminating or nonterminating, is calleda real number.

Periodic decimal fractions are called rational numbers. Every

rational number may be written in the form of a ratio, , of two

integers p and q y and vice versa.

Nonperiodic decimal fractions are called irrational numbers.

If X is a certain set of real numbers, then the notation x£Xmeans that the number x belongs to X, and the notation x^Xmeans that the number x does not belong to X.

A set of real numbers x satisfying the inequalities a < x < b ywhere a and b are fixed numbers, is called an open interval (a, b).A set of real numbers x satisfying the inequalities a^x^b is

called a closed interval [a, b]. A set of real numbers x, satisfyingthe inequalities a^x


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< x < + oo hold true for any real number x.

The interval {a Ð e, a + s) is called the e-neighbourhood of thenumber a.

The set of real numbers x > M is called the M -neighbourhoodof the improper point +00.

The set of real numbers x < M is called the M- neighbourhoodof t e improper point Ð-00.

The absolute value of a number x (denoted \x\) is a numberthat satisfies the conditions

\x\ = Ðx if x < 0;\x\ =x if xÔ.

12

Ch. I . Introduction to Mathematical Analysis

The properties of absolute values are:

(1) the inequality j a; | ^ a means that Ð a^xâ;

(2) the inequality |x|>a means that xâ or jc\\x\ Ð \y\\;

xy\=\x\\y\x

y

1.1.1. Prove that the number

0.1010010001... 1000... 01...

is irrational.

Solution, To prove this, it is necessary to ascertain that the


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given decimal fraction is not a periodic one. Indeed, there are nzeros between the nth and (n+\)th unities, which cannot occurin a periodic fraction.

1.1.2. Prove that any number, with zeros standing in all deci-mal places numbered 10" and only in these places, is irrational.

1.1.3. Prove that the sum of, or the difference between, a ra-tional number a and an irrational number P is an irrationalnumber.

Solution. Consider the sum of a and p. Suppose a + P = y is arational number, then P = Y Ð a is also a rational number, sinceit is the difference between two rational numbers, which contra-dicts the condition. Hence, the supposition is wrong and the numbera + P is irrational.

1.1.4. Prove that the product a|J and the quotient a/[i of arational number a ^= and an irrational number P is an irrationalnumber.

1.1.5. (a) Find all rational values of x at which y=\/ x* + x + 3

is a rational number.

Solution, (a) Suppose x and y=\/~x


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(b) Prove that \/~2 is an irrational number.1.1.6. Prove that the sum V~3 + V2 is an irrational number.Solution. Assume the contrary, i.e. that the number [/3 + |/*2is rational. Then the number

is also rational, since it is the quotient of two rational numbers.Whence the number

K2 = l[(K3 + K2)-(/3-K2)l

is rational, which contradicts the irrational nature of the numberV2 (see Problem 1.1.5). Hence, the supposition is wrong, and thenumber K3 + 1/2 i s irrational.

1.1.7. Prove that for every positive rational number r satisfyingthe condition r 2 < 2 one can always find a larger rational numberr + h(h>0) for which (r + hf < 2.

Solution. We may assume h0) for which (s Ð k) 2 > 2.

1.1.9. Solve the following inequalities:

(a) \ 2x Ð 3 | < 1;

(b) (x-2) 2 >4;

(c) x 2 + 2x Ð 8x 2 Ð 7x+ 12.

Solution, (a) The inequality |2# Ð 3|


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1.1.10. Find out whether the following equations have anysolutions:

(a) \x\=x + b] (b) |x|=x Ð 5?

Solution, (a) Atx^O we have x = x + 5. Hence, there are nosolutions. At x < we have Ð x=-x + 5, whence x== Ð 5/2. Thisvalue satisfies the initial equation.

(b) At x>0 we have x = x Ð 5. Hence, there are no solutions.At x < we have Ð x = x Ð 5, whence # = 5/2, which contradictsour supposition (#

at any values of x 9 the equality is satisfied at those values of x

at which 2x Ð 3>0, i.e. at *>3/2.

(b) The equality \a Ð 6| = |a| Ð \b\ holds true if and only if aand b have the same sign and

In our case the equality will hold true for the values of x atwhich

x x Ð 4>jc 2 + 2.Whence __


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x 2 Ð 2>1; \x\^V$-

1.1.13. Solve the inequalities:

(a) | 3* Ð 5| Ð | 2JC + 3 | > 0;

(b) \x 2 Ð 5x| > |jc 2 | Ð |5x|.

1.1.14. Find the roots of the following equations.

(a) | sin x | = sin x + 1;

(b) x 2 Ð 2\x\ Ð 3 = 0.

Solution, (a) This equation will hold true only for those valuesof x at which sin*(x), and soforth. The set of values of the function y(x) is called the rangeof the given function.

In particular, the functions defined by the set of natural num-bers 1, 2, 3, are called numerical sequences. They are written


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in the following way: x Xi x 2y ... or {x n \.

1.2.1. Given the function / (x) =(*+ \)/(xÐ 1). Find f(2x), 2f(x),fix 2 ), [f(x))>.

Solution.

f(20=§±i; 2f W _2£±};

/-£!; tfW]'-(f±{)'.

1.2.2. (a) Given the function

Show that at x lt x 2 £(Ð 1, 1) the following identity holds true:

/(*>+/(*)=/ (t££-).

16

Ch. I. Introduction to Mathematical Analysis

Solution. At 1, 1) we have (1 Ð x)/(l + x) > and hence

f(x l)+ f M = log ^ + log = log ;; ;g . ( i,

On the other hand,

^1 + ^2

^ ^ *l + *2 \ __ j Q g 1 ~f- X X X 2 _ j Q g 1 Ð X x Ð X 2

\l-jrX1X2J j -^l ~l~ ^ 2 1 -\~ x i x 2~i~~ x i ~h x 2

1 -j- XiX 2

(1Ð *i) (\Ðx 2 ]

log (!+*!) (l+x 2 ) '

which coincides with the right-hand member of expression (1).(b) Given the function / (x) = (a x + a' x )/2 (a > 0). Show that

f{x + y) + f(x-y) = 2f(x)f{y).

1.2.3. Given the function f (x) = (x+ l)/(x»Ð 1). Find /(Ð 1);f(a+\); f(a)+\.

1.2.4. Given the function f(x) = x 3 Ð 1. Find


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/(»>-/("> ( ^ a) and ^«±*.

6 Ð a

1.2.5. Given the function

( 3-^Ð1, Ð\


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then

S (X) Ð S^ AMN Ð X 2 -

If x> V2 y thenS(x) =4 Ð (2 K2~Ð -^) 2 = Ð jc 2 +

+4* j/y

4.

Thus,

S(*) =

Fig. 1

0


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18


Determine the coefficients a, b y c from the above system. We have:a = 3; b=r.Ð2; c = 5\ hence / (x) = 3x 2 Ð 2x + 5.

1.2.11. Find a function of the form

f(x) = a + bc* (c>0),if /(0)=15; f(2) = 30; f(4) = 90.

1.2.12. Find cp [xp (a:)] and [cp (jc)] if

cp (jc) = x 2 and ^(x) = 2 x .

Solution.

i|)[


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\f(x) +


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This inequality will be satisfied if

-5^1>1, or x 2 Ð 5x + 40 at any x, the problem is reduced to solvingthe inequality

4 + 2 sin x ^

Whence

3-Ð 7 2 .By solving the latter inequality we obtain

_£ + 2£:ri x. This inequality is satisfied at x < 0.Hence, the function is defined in the interval (Ð oo, 0).

1.2.22. Find the domains of definition of the following functions:

(a) / (x) = j/arc sin (log 2 x) ;

(b) / (x) = log 2 log 3 log 4 x;

(C) / W = l + 2 a-in, + _J_.

(d) f (x) = log|4-x 2 |;

(e) f (x) = V cos (sin x) +- arc sin ~£ x .Find the ranges of the following functions:

( f ) & = 2- cos 3* ;

(g) y=j^ ·

Solution, (a) For the function /(x) to be defined the following


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inequality must be satisfied

arc sin (log 2 x) ^ 0,

whence ^ log 2 x^ 1 and l^x^2.

(b) The function log 2 log 3 log 4 x is defined for log 3 log 4 x>0,whence log 4 x>l and x>4. Hence, the domain of definition isthe interval 42,

but the inequalities Ð l^x^l and x>2 are incompatible, thatis why the function is not defined for any value of x.

§ 1.2. Function. Domain of Definition

21

(e) The following inequalities must be satisfied simultaneously:

1 + x 2

cos (sin x)^0 and

2x

< 1.

The first inequality is satisfied for all values of x, the second,for |*|=1. Hence, the domain of definition of the given functionconsists only of two points x=±\.

(f) We have

cos 3x ~ Ð Ð -

Since


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-1 < cos3x< 1, we have Ð Ð -< 1,

whence, taking into account that y > 0, we obtainÐ y^2yÐ or y 1.

(g) Solving with respect to x y we obtain

x - Ty 'The range of the function y will be determined from the relation

1 Ð 4*/ 2 >0.

Whence

1.2.23. Solve the equation

arc tanVx(x + 1)+ arc sin l^x 2 + x+ 1 = jx/2.

Solution. Let us investigate the domain of definition of the func-tion on the left side of the equation. This function will be defined

for

x 2 + *>0, 0< x 2 + x+ 1 < 1,

whence x 2 -\-x = 0.

Thus, the left member of the equation attains real values onlyat x t =0 and # 2 = Ð 1. By a direct check we ascertain that theyare the roots of the given equation.

This problem shows that a study of domains of definition of afunction facilitates the solution of equations, inequalities, etc.

1.2.24. Find the domains of definition of the following functions:

, x 2xÐ 3

(a) y= r « _ ;

(b) y = \og sin (a:Ð 3) + j/l6 Ð x\

22


(c) y = V 3 Ð x + arc cosÐ ^Ð ;

1.2.25. The function f (x) is defined on the interval [0, 1]. Whatare the domains of definition of the following functions:


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(a) /(3r>); (b) f(x-5); (c) f(tan*)?

Solution. The given functions are functions of functions, or su-perpositions of functions, i. e. composite functions.

a) Let us introduce an intermediate argument u = 3x 2 . Then thefunction f(3x 2 )j=f(u) is defined if 0


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3. The zeros of the function.

4. The sign of the function in the intervals between the zeros.

5. Is the function bounded and what are its minimum and ma-ximum values?

The above items do not exhaust the analysis of a function, andlater on their scope will be increased.

1.3.1. Find the intervals of increase and decrease of the func-tion / (x) = ax 2 -f- bx + c y and its minimum and maximum values.

Solution. Isolating a perfect square from the square trinomial,we have

If a > 0, then the function f(x) will increase at those values of xsatisfying the inequality x + b/(2a) > 0, i. e. at x > Ð b/(2a), anddecrease when x + b/(2a) < 0, i.e. at x 0,

the function f (x) decreases in the interval ( Ð oo, Ð Ð ) and inc-

reases in the interval ( Ð b/(2a), +oo). Obviously, at x = Ð b/(2a)the function / (x) assumes the minimum value

At a > the function has no maximum value.

Similarly, at a < the function f (x) will increase in the inter-val ^ Ð oo, Ð ^ and decrease in the interval ( Ð b/(2a), oo); at

x = Ð b/(2a) the function f (x) takes on the maximum value

£ r / b \ 4ac Ð b*

(b) Find the rectangle with the maximum area from among allrectangles of a given perimeter.

Solution, (a) Apply the results of Problem 1.3.1: a Ð 3 > 0, b=5,c = Ð 1. The minimum value is attained by the function at thepoint x = Ð 5/6

whereas it has no minimum value.

1.3.2. (a) Find the minimum value of the function


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y = 3x 2 + 5x Ð 1.

4ac Ð b*4a

3712'

(b) We denote by 2p the length of the perimeter of the requiredrectangle, and by x the length of one of its sides; then the area 5

24

Ch. J. Introduction to Mathematical Analysis

of the rectangle will be expressed as

S = x(p Ð x) or S = px Ð x 2 .

Thus, the problem is reduced to the determination of the maximumvalue of the function S (x) = Ð x 2 + px. Apply the results of Prob-lem 1.3.1: a = Ð 1 < 0, b = p, c = 0. The maximum value is attai-ned by the function at the point x = Ð b/(2a)=p/2. Hence, one

of the sides of the desired rectangle is p/2, the other side beingequal to p Ð x = p/2, i.e. the required rectangle is a square.

1.3.3. Show that

(a) the function f (x) Ð x 3 + 3x + 5 increases in the entire domainof its definition;

(b) the function g(x) = x/(l + x 2 ) decreases in the interval (1, +oo).Solution. The function is defined for all points of the number

scale. Let us take arbitrary points x x and x 29 x x < x 2 on the numberscale and write the following difference:

Since x 2 Ð x x > and the expression in the brackets is positiveat all x 1 and jc 2 , then / (x 2 ) Ð f (xj > 0, i.e. / (* 2 ) >f(Xj), whichmeans that the function f (x) increases for all values of x.

1.3.4. Find the intervals of increase and decrease for the follo-wing functions:


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(a) / (x) = sin #-|-cos x;

(b) tan(*+n/3).

Solution, (a) Using the familiar trigonometric formulas, we find

It is known that the function cos a: decreases in the intervals

and increases in the intervals

(2nÐ\)ji 0).

§ 1.3. Investigation of Functions

25

Solution. The given function can be represented as:

f (x) = Vcl 1 -V b 2 cos (x Ð a),

where cos a = a/)/ a 2 + b 2 , sin a = 6/ Ka 2 + b'\ Since |cos(x Ð a)|^ 1,the maximum value of f (x) equals +\/ r a 2 + t) z (at cos(a: Ð a)-=l),the minimum value of f (x) being equal to j/a 2 + fr 2 (atcos (xÐa) = Ð 1).


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1.3.6. Find the minimum value of the function

f(x) =3


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l+x'

(c) f(x) = 2x s Ð x+l;

(d) /W = *S^T-

Solution, (a) It can be seen that /(+ x) + f (Ð x) ^=0. Indeed,

/(+ x) + f(- ^)-log(^ + KTT^ 2 ) + log(-^+KTT^ 2 ) =

= log ( 1 + * 2 -- x 2 )=0,

hence, / (x) = Ð / ( Ð *) for all a:, which means that the function is odd.

(b) m_ ^ = io g = log v 1 = Ð io g 1

\+x) 6 1+jc'

Thus, f(Ðx) = Ð f(x) for all x from the domain of definition(Ð 1, 1). Hence, the function is odd.

1.3.10. Which of the following functions is (are) even and whichis (are) odd?

(a) f(x) = 4 Ð 2x* + sm 2 x\

(b) f(x) = V \+x + x 2 Ð \ r \Ð x + x 2 ;

(d) f (x) = sin x + cos x;

(e) f (x) = const.

1.3.11. Prove that if f (x) is a periodic function with period T 9then the function / (ax + b), where a > 0, is periodic with period T/a.

Solution. Firstly,

f[a(x+ T/a) + b]=f [(ax + b) + T] = f (ax + b) y

since T is the period of the function f(x). Secondly, let 7\ be apositive number such that

f[a(x+T 1 ) + b]=f(ax + b).

Let us take an arbitrary point x from the domain of definitionof the function f (x) and put x' = (# Ð b)/a. Then

/ + = / (ai=^+ ft) =/(x) = f [a {x' + T l ) + b] =

= f(ax' + b + aT l ) = f(x + aT l ).

Whence it follows that the period T7> andT/a is the period of the function f(ax + b).

Note. The periodic function / (x) = A sin (ayx + cp), where /I, co, 9are constants, is called a harmonic with amplitude |j4|, frequency co


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§ 1.3. Investigation of Functions

27

and initial phase (p. Since the function sin* has a period 2k, thefunction A sin(o)A: + (p) has a period T = 2n/(s>.

1.3.12. Indicate the amplitude \A\, frequency co, initial phase (pand period T of the following harmonics:

(a) / (x) = 5 sin \x\

(b) /(*) = 4sin(3* + n/4);

(c) /(x) = 3sin(jt/2) + 4cos(x/2).

1.3.13. Find the period for each of the following functions:

(a) / (a;) = tan 2x;

(b) f(x)=cot(*/2);

(c) \(x)^s\x\ 2nx.

Solution, (a) Since the function tanx has a period Jt, the functiontan 2x has a period n/2.


(a) f (x) = s\n* x + cos 4 x;

(b) /(a:) = |cosa:|.

Solution, (a) sin 4 x + cos 4 x= (sin 2 x + cos 2 x) 2 Ð 2 sin 2 x cos 2 x =

= 1Ð y sin 2 2x = 1 Ð cos 4x) = -| + -j-sin ^4x + y) ;

whence T = 2n/a = 2n/4 = n/2 .

(b) /(x) = |cosjc| = Kcos 2 x^K(1 + cos2x)/2; but the functioncos 2.x; has a period T = n\ hence, the given function has the sameperiod.

1.3.15. Prove that the function / (x) = cos x 2 is not a periodic one.

Solution. Let us prove the contrary. Suppose the function has a

period T\ then the identity cos (x + T) 2 = cos x 2 is valid.

By the conditions of equality of cosines for a certain integer kwe have

x 2 + 2Tx + T 2 ±x 2 = 2nk.

But this identity is impossible, since k may attain only integral


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values, and the left member contains a linear or quadratic functionof the continuous argument x.

1.3.16. Find the greatest value of the function

f(x)= r 2 = .

1.3.17. Which of the following functions are even, and which areodd:

(a) /W=^(l-x) 2 + 3 /(l+x) 2 ;

28


(b) f(x) = x*-\x\;

(c) / (x) = x sin 2 x Ð x 3 ;

(d) /(x) = (l + 2*) 2 /2*?


(a) / (x) = arc tan (tan x)\

(b) f (jc)-2cos^.

1.3.19. Prove that the functions

(a) f(x) = x + s\nx; (b) f (x) = cos Vx

are non-periodic.

§ 1.4. Inverse Functions

Let the function y = f(x) be defined on the set X and have arange V\ If for each y£Y there exists a single value of x such thatf(x)=y y then this correspondence defines a certain function x Ð g(y)called inverse with respect to the given function y = f(x). The suf-ficient condition for the existence of an inverse function is a strictmonotony of the original function y = f{x). If the function increases(decreases), then the inverse function also increases (decreases).

The graph of the inverse function x = g(y) coincides with that of

the function y = f(x) if the independent variable is marked off alongthe (/-axis. If the independent variable is laid off along the x-axis,i.e. if the inverse function is written in the form y = g(x) y thenthe graph of the inverse function will be symmetric to that of thefunction y = / (x) with respect to the bisector of the first and thirdquadrants.

1.4.1. Find the inverse to the function y = 3x + 5.

Solution. The function y = Sx + 5 is defined and increases through-


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out the number scale. Hence, an inverse function exists and in-creases. Solving the equation y = 3x + 5 with respect to x we obtainx = (y-5)/3.

1.4.2. Show that the function y = k/x (k=^=0) is inverse to itself.Solution. The function is defined and monotonic throughout the

entire number scale except x = 0. Hence, an inverse function exists.The range of the function is the entire number scale, except # = 0.Solving the equation y = k/x with respect to x, we get x = k/y.

1.4.3. Find the inverse of the function

y = \°ga(x + V*+l) 9 (fl>0, a¥= 1).Solution. The function y = log a (x + }/~x 2 + l) is defined for all x,since ]fx 2 + 1 > \x\ y and is odd [see Problem 1.3.9 (a)]. It increases

§ 1.4. Inverse Functions

29

for positive values of x, hence, it increases everywhere and has aninverse function. Solving the equation

y=\og a (x + Vlfi+l)

with respect to x, we find

a? = x + V x 2 + l; a-y= Ð x+Vx 2 + 1,

whence

x = y (a^ Ð = sinh (*/ lna).

1.4.4. Show that the functions

f(x)=x 2 Ð x + 1, x^ 1/2 and cp (*) = 1/2 + j/x Ð 3/4

are mutually inverse, and solve the equation

x* Ð x + 1 = 1/2+ Vx Ð 3/4.

Solution. The function # = a; 2 Ð x + 1 = (xÐ 1/2) 2 + 3/4 increasesin the interval 1/2^a:


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Let us now solve the equation

x* Ð x+l = l/2 + Vx Ð 3/4.

Since the graphs of the original and inverse functions can intersectonly on the straight line y = x, solving the equation x 2 Ð x+l = xwe find x = 1 .

1.4.5. Find the inverse of y = smx.

Solution. The domain of definition of the function y = s\nx isthe entire number scale, the range of the function is the interval[ Ð 1, 1]. But the condition of existence of an inverse function isnot fulfilled.

Divide the x-axis into intervals tin Ð n/2^x^nn + n/2. If n iseven, then the function increases on the intervals tin Ð n/2^x^^nn + n/2; if n is odd, the function decreases on the intervalsnn Ð ji/2^ x^nn + n/2. Hence, on each of the indicated intervalsthere exists an inverse function defined on the interval [Ð1, 1].

30


In particular, for an interyal Ð jt/2 ^ ji/2 there exists an in-verse function jt = arcsinr/.

The inverse of the function y = s\x\x on the interval mi Ð^*^wt+jt/2 is expressed through atcsin y in the following way:

x = (Ð l) B arcsin# + wr (ai = 0, +1, ±2, ...).

1.4.6. Find the inverse of the given functions:

(a) y = s\n(3xÐ 1) at Ð (jt/6+ l/3)< (jt/6+ 1/3);

(b) # = arcsin(*/3) at Ð 3


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(c) f(x) = s\n 2 x Ð 2sinx;

(d) / (x) = arc cos (cos x);

(e) f (x) = Vsm x;

(f) f( X ) = X x ' l °*x.

Solution, (a) The domain of definition of the function/^) is theentire number scale. The function f (x) is even, hence its graph issymmetrical about the ordinate axis and it issufficient to investigate the function at x^O.

Let us single out a perfect square f(x) == (x 2 Ð 1) 2 + 2. Since the first summand(x 2 Ð 1) 2 ^0, the minimum value of the func-tion, equal to 2, is attained at the pointsx = ±l (see Fig. 2).

The function f (x) decreases from 3 to 2 onthe closed interval O^x^l and increasesunboundedly on the open interval 1 < x < oo.

(b) The domain of definition of the func-tion f (x) is the entire number scale. The fun-ction / (x) is odd, therefore its graph is symmetrical about the originof coordinates and it is sufficient to investigate the function atx ^0.

Since /(0) = 0, the graph passes through the origin. It is obviousthat there are no other points of intersection with the coordinate

Fig. 2

§ 1.5. Graphical Representation of Functions

31

axes. Note that |/(jc)|0or l+x 2 >2|*|,whence

Since / (x) > at x > and f (1) = 1, in the interval [0, oo) themaximum value of the function f (x) equals 1, the minimum valuebeing zero (see Fig. 3).

1 ]


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I I 1

/ i i i

-3 "2 -1 /

12 3'

-/

Fig. 3

Let us prove that the function increases on the closed interval 0^


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32


(d) The domain of definition of the function is the entire numberscale. Indeed, |cosx|^l at any x y hence, arc cos (cos*) has ameaning. The function f (x) is a periodic one with the period 2n yhence, it is sufficient to sketch its graph on the interval [0, 2jx].

But on this interval the followingequality is true:

f(x)= l *. 0


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X

Fig. 5

the function is not defined in the whole interval [0, 2ji], it isdefined only in the interval [0, ji], as in the interval (jx, 2n) theradicand is negative. The graph is symmetrical about the straight

Fig. 6

line a: = ji/2, as well as the graph y = s'mx (see Fig. 6). Here wehave an example of a periodic function which does not exist inthe infinite set of intervals.

(f) The domain of definition of the function is

< x < 1 and 1 < x < oo.

§ 1.5. Graphical Representation of Functions

33

Reduce the formula to the form

f(x) = x xn °* x = x Xo ** 10 = jo.

Hence, the graph of the given function is the half-line #=10in the right-hand half plane with the point x = 1 removed (see Fig. 7).

1.5.2. Sketch the graphs of functions defi-ned by different formulas in different inter-vals (and in those reducible to them):sin x at Ð jx


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Ð 2 at x>0,1/2 at x= 0,

Ð x 3 at x < 0;

(c) y=x + V#; _

(d) y=2/(x+\f?).

(a) */=<

(b)0=

0< x< 1,1 < a; < 4;

Fig. 7

Solution, (a) The domain of definition of the function is theinterval [ Ð it, 4]. The graph of the function consists of a portionof the sinusoid y = s\x\x on the interval Ð Jt< x < 0, straight line

9,

2

1


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I i .1 1Ð

12 3 4

-i

1/2 {

-1

O 1

-1

-2

Fig. 8

Fig. 9

i/ = 2 on the interval (0, 1] and a part of the branch of the hyper-bola y=\l{xÐ\) on the interval (1,4] (see Fig. 8).

(b) The graph of the function consists of a portion of a cubicparabola, an isolated point and a half-line (see Fig. 9).

(c) The function may be given by two formulas:


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I 2*. if x>0,y== \ 0, if x 0). Thus, the graphof our function is the right-hand part of an equilateral hyperbola(see Fig. 11).

y

1

Fig. 11

1.5.3. Sketch the graphs of the following functions:

(a) ^ = cosx+|cosx|;

(b) y=\x + 2\x.

\ 2cosjc at cos* ^5 0,Solution, (a) cosa; + | cos*| = < q at COSA;


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Fig. 12

the points where cosx


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\ '2/Ð ' \ / 1 >

-1

1 V/ 3

-1

Fig. 13 Fig. 14

Finally, at x


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Solution. Being the product of two odd functions y x = x and*/ 2 = sin*, the function y is an even one, that is why we shallanalyse it for x^O.

We draw graphs for y x =x and y 2 =smx (the broken lines inFig. 16).

At the points where r/ 2 = sin# = 0, y = y 1 .y 2 = i and at thepoints where y 2 = smx = ± 1, y=± yi = ±x. The latter equality

x

Fig. 16

indicates the expedience of graphing the auxiliary function y 3 = Ð x.

Marking the indicated points and joining them into a smoothcurve, we obtain the required graph (the solid line in Fig. 16).

1.5.7. Sketch the graph of the function y = x(x 2 Ð 1) by multi-plying the ordinates of the graphs y t =x and y 2 z=x 2 Ð 1.

1.5.8. Graph the following functions:(a) y = x/(x 2 Ð 4), (b) y = 1 /arc cos x.

Solution, (a) Since the function is odd, it is sufficient to inves-tigate it for x ^ 0.Let us consider it as the quotient of the two functions:

y x = x and y 2 = x 2 Ð4.

Since at x = 2 the denominator y 2 = 0, the function is not de-fined at the point 2. In the interval [0, 2) the function y x increasesfrom to 2, the function y 2 is negative and|r/ 2 | = 4 Ð x 2 decreasesfrom 4 to 0; hence, the quotient f(x)=yjy 2 is negative and in-creases in absolute value, i.e. f(x) decreases in the interval [0,2)from to Ð oo.

In the interval (2, oo) both functions are positive and increasing.Their quotient decreases since from 2^x t


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37

The indicated quotient tends to zero as x Ð ► oo, since y = 1 _^ X 2 ~^ 0-

The general outline of the graph is presented in Fig. 17 (threesolid lines).

(b) Denote y 1 = arc cos x. The domain of definition of this func-tion | x At x=\ we have y l = 0, hence, y=\/y l Ð * oo at

Fig. 17 Fig. 18

xÐ + 1, i. e. x= 1 is a vertical asymptote. The function y l decreaseson the entire interval of definition [ Ð 1, 1), hence y=\/y l incre-ases. The maximum value y 1 = n is attained at *--=Ð 1. Accor-dingly, the minimum value of the function is 1/jx. The solid line

in Fig. 18 represents the general outline of the graph.

Simple Transformations of Graphs

I. The graph of the function y = f(x + a) is obtained from thegraph of the function y = f(x) by translating the latter graph alongthe x-axis by | a | scale units in the direction opposite to the signof a (see Fig. 19).

II. The graph y = f(x) + b is obtained from the graph of thefunction y = f(x) by translating the latter graph along the y-ax\sby \b\ scale units in the direction opposite to the sign of b (seeFig. 20).

III. The graph of the function y = f(kx)(k>0) is obtained fromthe graph of the function y = f (x) by "compressing" the latter graphagainst the *y-axis in the horizontal direction k times at k > 1 andby "stretching" it in the horizontal direction from the r/-axis \\ktimes at k < 1 (see Fig. 21).

38

Ch. J. Introduction to Mathematical Analysis

IV. The graph of the function y = kf (x) (k > 0) is obtained fromthe graph of function y = f(x) by "stretching" it in the horizontaldirection k times at k > 1 and "compressing" it against the #-axis(i. e. vertically) \/k times at k < 1 (see Fig. 21).


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u>-0 / vr

,o yuf(x)

y=f(x)+l,l0 thegraph of the function y = f(x) is retained, then this retained partof the graph is reflected symmetrically about the #-axis, thus de-termining the graph of the function for jc


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1.5.9. Graph the function

» = 3J/

-2(a; + 2.5) Ð 0.8

by transforming the graph y = ]fx.

Solution. Sketch the graph of the functiony=z\fx (which is the upper branch of theparabola y 2 = x) (Fig. 24, a), and transformit in the following sequence.

Sketch the graph of the function y = 3]/2x Fig. 23

by enlarging 3 V 2 times the ordinates of

the points on the graph of the function y= \ x and leaving theirabscissas unchanged (see Fig. 24, b).

Then sketch the graph of the function y = 3V Ð 2x which will

be the mirror image of the preceding graph about the (/-axis (seeFig. 24, c).

Fig. 24

By shifting the obtained graph 2.5 scale units leftward and then0.8 u nit downwar d draw the desired graph of the function

y=3VÐ 2(jc + 2.5)Ð 0.8 (see Fig. 24, d).

1.5.10. Graph the function # = 3cos# Ð j/3sin;t by transformingthe cosine curve.

40


Solution. Transform the given functiony = 3 cos xÐ V 3 si n x = 2 j/3~(

Q ' j-cosx Ð y sin jc


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= 2 K3 cos

Thus, we have to sketch the graph of the function# = 2VTcos (x + ji/6),

which is the graph of the function */ = 2 K 3cosjc translated by ji/6leftward. The function has a period of 2ji, hence it is sufficient to

draw its graph for Ð n ^Zx ^ ji

ZVT

/ // /

/' 1

1 //

\\ \

\ \

\\ 1 >

/ fa/2

I l

>^ -2V3

f\\7t/2 %

\\

·X

Fig. 25

Fig. 26


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The graph of any function of the form */ = acosx + frsin x, wherea and b are constants, is sketched in a similar way.

1.5.11. Graph the following functions:(a) yÐ^i

< b > f-JTZ^

(c) ^ x 2 + a:+ 1, ifÐ l


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1.5.12. The function y = f(x) is given graphically (Fig. 26).Sketch the graphs of the following functions:

(a) y = f(x+ 1);

(b) y = f(x/2);

(c) y=\f(x)\;

(d) » = (|/(*)|±/(*))/2;

(e) y = \f(x)\/f(x).

§ 1.6. Number Sequences. Limit of a Sequence

The number a is called the limit of a sequence x u x 2y x n ,... as n Ð ►oo, aÐ lim x n if for any e>0 there exists a number

n -+ co

N(e)>0 such that the inequality |jt¹ Ð a\ Af(e).

A sequence which has a finite limit is said to be convergent.A sequence {x n } is called infinitely small if lim.*^ Ð 0, and infi-nitely large if lim^^oo.

1.6.1. Given the general term of the sequence {x n }:

sin (nn/2)

Write the first five terms of this sequence.

Solution. Putting consecutively n=\ 9 2, 3, 4, 5 in the general

term x ni we obtain

sin (jt/2) « #

l »

1

3 ~~ 3

sin (4it/2) ~

: 4 "" U;

sin(5jt/2) _ 1

5 ~~ 5 *


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42


1.6.2. Knowing the first several terms of the sequence, write oneof the possible expressions for the general term:

(3 ' I' 8 » 13 » 18 ' 23 ;

(b) 1, -g-, 2, -g- , 3, , 4, -g-.

Afofe. A knowledge of the first several terms of a sequence is notsufficient to define this sequence. That is why this problem shouldbe understood as one of finding a certain simple inductive regula-rity compatible with the given terms.

Solution, (a) Note that the numerator of each of the given termsof the sequence equals the square of the number of this term plus

unity, i.e. n 2 +\, while the denominators form the arithmetic prog-ression 3, 8, 13, 18, ... with the first term ^=3 and the com-mon difference d = 5. Hence,

a¹ = a 1 + d(nÐ 1 ) = 3 H- 5 (az Ð l) = 5n Ð 2,thus we have

_ n* + l*»~~5/i Ð 2 '

(b) Here the general term of the sequence can be written withthe aid of two formulas: one for the terms standing in odd places,the other for those in even places:

j k at n = 2kÐl 9

x n = \ 1/(6+1) at n = 2k.

It is also possible to express the general term by one formula,which will be more complicated, for instance,

^=^[i-(-i)i+^- 2 [i+(-i)"]-

1.6.3. Find the first several terms of the sequence if the generalterm is given by one of the following formulas:

(a) jc¹ = sin (nn/3)\

(b) x n = 2~ n cos nn\

(c) x n =(l + l/n)\

1.6.4. Using the definition of the limit of a sequence, prove that

(a) lim x n =l if x n = (2nÐ l)/(2n+ 1),


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(b) limx n = 3/5 if x n = (3n 2 + l)/(5n 2 Ð 1). Beginning with whichn is the inequality | Ð 3/5 | < 0.01 fulfilled?

Solution, (a) For any e > let us try to find a natural numberAf(e) such that for any natural number n>N(z) the inequality\x n Ð l| 1/eÐ V 2 - Hence the integral part of the number 1/e Ð V, maybe taken as jV(e), i.e. W = £(1/8 Ð 7,).

So, for each e > we can find a number A/ such that from theinequality n > N it will follow that \x n Ð 1 1 < e, which means that

lim f^ = l.

(b) Let us find the absolute value of the difference Ð 3/5|:3/z 2 +l 3 I 8

5/z 2 Ð 1 5 | 5(5/i 2 Ð 1)'Let e > be given. Choose n so that the inequality

5(5/i 2 -l)

is fulfilled.Solving this inequality, we find

_ 8 . 1 .1 t/8 + 5b


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Putting

N = £ (i/?±5),

we conclude that at n> N

K-3/5|


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quence for which this distance exceeds 0.001, i.e.

19

3(9/z + 4) ^ 1 000

whence

1


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than the constant number 1 / 2 ; hence, there exists such e > 0, say,e= 1 / 29 that the inequality

2ri l Ð 9

Ð

>

1

holds true for any n^3.

The obtained inequality proves that / = is not the limit of the

given sequence.

1.6.7. Prove that the sequence

112 1

1, ^

with the general term

\/n, if /2-2/eÐ 1,

n/(n + 2), if n = 2k,has no limit.

Solution. It is easy to show that the points x n with odd num-bers concentrate about the point 0, and the points x n with evennumbers, about the point 1. Hence, any neighbourhood of the point 0,as well as any neighbourhood of the point 1, contains an infiniteset of points x n . Let a be an arbitrary real number. We can always

choose such a small e > that the e-neighbourhood of the point a will


45


52/692

not contain at least a certain neighbourhood of either point orpoint 1. Then an infinite set of numbers x n will be found outsidethis neighbourhood, and that is why one cannot assert that all thenumbers x m beginning with a certain one, will enter the e-neigh-bourhood of the number a. This means, by definition, that thenumber a is not the limit of the given sequence. But a is an arbit-rary number, hence no number is the limit of this sequence.

1.6.8. Prove that lim* n =l if x n = (3 n + l)/3 n .

1.6.9. Prove that limx¹ = 2 if x n = (2n + 3)/(n + 1). Find thenumber of the term beginning with which the inequality| (2n + 3)/(/2+ 1) Ð 2 | < e, where 8 = 0.1; 0.01; 0.001, is fulfilled.

1.6.10. Prove that the sequenceJ_

2

with the general term

1 _L i. _L L

2 ' 2 ' 4 ' 4 ' 8

I l -^rm if n is odd **»=f i .

Ð j77 if n is even,

{ 2 n/2

has no limit.

1.6.11. Prove that at any arbitrarily large a>0 limA;¹ = ifx n = a n /nl

Solution. Let a natural number k > 2a. Then at n > k

a n a a a f a a a \ ( a a a x ^

n\~ \ 2 n \ 1 2 k ) \k+l k + 2 n


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Since lim(l/2)" = (prove it!), then at a sufficiently large n wehave: (4-)


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1.6.15. Show that the sequence with the general termx n = ( Ð l) n 2/(5f/n+l) is infinitely small as nÐ>oo. Find a num-ber N beginning with which the points x n belong to the interval(-1/10, 1/10).Solution. Take an arbitrary e>0 and estimate \x n \:

I l- 2 ^ 2 ^ 2 - 1

l JC »l"" c 3/ . f ^ - 3/- 3/- 3/-'

That is why |xj 1/e 3 . Hence limx n = 0, i.e.

n oo

the sequence is infinitely small.

We take now 8=1/10. Since |^ n |< l/\/n y x n will necessarilybe smaller than 1/10 if \lV~n< 1/10 or n > 1000. Hence N maybe taken equal to 1 000. But we can obtain a more accurate resultby solving the inequality

\ X n\ = 3/- < TO ·

It holds true at n > (19/5) 3 = 3. 8 3 = 54.872. Hence N may be

taken equal to 54^1000.

1.6.16. It is known that if x n = a + a n , where a n is an infinite-simal as nÐ ^00, then lim x n = a. Taking advantage of this rule,

n -»· oo

find the limits:

3"+! + sin (wi/4) 2* + (Ð 1)"


47

c < . . 3 w+1 -f- sin (nn/4) , , sin (nn/A) .

Solution, (a) x n = "s^ Ð Ð 3 + where a n = Ð v 3 ¹ is

an infinitesimal as Ð ► oo, hence lim x n = 3.

n -+ oo

1.6.17. Prove that lim y^n=l.

n -+

Solution. Let us prove that the variable Y n can be representedas the sum l+a¹, where a n is an infinitesimal as n Ð ► oo.


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Let us put \/n= 1 + a n . Raising to the nth power we obtain

rc= (1 +aj n = 1 +Aia¹ + ^=^a^+ . . . +a" ny

wherefrom we arrive at the conclusion that for any n > 1 the fol-lowing inequality holds true:

n>l + njn^l) a%

(since all the terms on the right are non-negative). Transposing theunity to the left and reducing the inequality by n Ð 1 we obtain

whence it follows that 2/n > a 2 n or > a n > 0. Since

lim V r 2/n = i lim a n also equals zero, i. e. a n is an infinitesimal.

n -* oo n oo

Hence it follows that

lim yn=l.

n -* oo

1.6.18. Prove that the sequence with the general term

is infinitely large as nÐ >oo.

Solution. Let us take an arbitrary positive number M and solvethe inequality

z V n > AJ.

Taking the logarithm, we obtain

Vn> log, /W, n> (log 3 M) 3 .

If we now take /V = E (log 3 A4) 3 , then for all n> N the inequa-lity |x n |>A4 will be fulfilled, which means that the sequence isinfinitely large.

1.6.19. Prove that

lim y/a = 1 (a > 0).

n -*■ oo

48


§ 1.7. Evaluation of Limits of Sequences

If the sequences {x n \ and {y n \ are convergent, then


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(1) Km (x n ± y n ) = liin x n ± lim

(2) lim - lim x n lim

If x¹ ^ */¹ then lim x n ^ lim y n .1.7.1. Find lim x¹ if

n qo

^ v 3n 2 +5n + 4 , 5b» + 2«»-3/i + 7 .

W x n~ 2 +n 2 ' W n ~ 4n 3 -2n+ll '

4n 2 -4n + 3 . l»+2»+ ...+«« .

Ð 2/i3 + 3/i + 4' w n_ 5/i3 + n+l

(e) * n = 1+2 +r + " -

3+A+l

Solution, (a) n

S+ 1

~ 2_n*

lim (3 + 5//z + 4//z 2 )lim x = Ð 7-o x Ð -3.

(d) Recall that

(2/i+l)

Hence

l* + 2 2 + 3 2 +... +n 2

2 + A-f-l

/i(/i+1)(2/i+1)_ 2n 3 + 3n 2 + Ai _ ^ n^ n 2

6(5/i* + /i+1) 6(5n3 + n+l) 30 _ L ^6 6»


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1 n 1 u 3

lim #¹ = 1/15.

1.7.2. Find lim x ny if

ft -+ 00

, , / 3n 2 + nÐ 2 \ 3 . / 2n 3 + 2n 2 +l \«

W Ð ^4¹2 + 2n + 7j ' W U" 3 + 7 " 2 + 3n + 4j ·

(c) x n = ybn; (d) x n = p i*;

{t)x n =yn>; (l)x n =^6n + 3.

Solution, (a) Urn (^^r)' =

_ ,. /' 3n 2 + n Ð 2 \ / 3n 2 + n Ð 2 \ / 3n 2 + n Ð 2 \ _«i m « U« 2 + 2n+7j U« 2 + 2« + 7j U/t 2 + 2/ l + 7j _

= lira

3+1/rt Ð 2//i 2 \ 3

¹ 4+2/rt + 7/rc 2

' 3_\ 3 _27,4 J -64

§ 17. Evaluation of Limits of Sequences

49

(c) In solving this example, and also the rest of the examplesof Problem 1.7.2, take advantage of the following equalities (see

Problems 1.6.17 and 1.6.19):

lim i/n=l and lim \/a=\. (1)

We have

lim x n = lim i/5n = lim lim \/n,

n co n -+ n -*■ co n -*■ co


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but from (1) it follows that limv^5=l and lim y^/i = 1 ; hencelim x n = 1 · 1 = 1 .

1.7.3. Find

' 2ft 3 . 1Ð 5ft 2 \

Solution. Summing the fractions, we obtain

2ft 3 Ð 13ft 2 + 3x nÐ i Oft 3 + 2ft 2 + 15ft + 3 *

Whence

_ r 2ft 3 Ð 13ft 2 + 3 _ 1lim x n - iirn^ 10n3+ 2 n 2 + i5 n + 3 - 5 ·

n -+ oo

Note. If we put

_ 2ft 3 _ 1Ð 5ft 2

Ð 2ft 2 + 3 ; ^ Ð 5ft+ 1 '

then the limit of their sum \im{y n + z n ) = 1/5, though each of thesummands is an infinitely large quantity. Thus, from the conver-gence of a sum of sequences it does not, generally speaking, followthat the summands converge too.

1.7.4. Find lim x n if

n -*■ co

(a) x ¹ = V2n+3 Ð VriÐl' 9

(b) x n = Vn 2 + n+ \ Ð]/ n n 2 Ð n+ 1;

(c) x n = n 2 (nÐVn 2 + \y,

(d) x n =¥n 2 Ð n 3 + n\

_ ]Tn*+\ + Vn .(ej x n Ð 4 / , Ð ,

( , 1Ð2 + 3 Ð 4 + 5 Ð 6+ ... Ð2ft

(h) x n = ± + ± + ±+...+.

1-2 1 2-3 1 3-4 1 ' ' * 1 ft(ft+l)

50


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Solution, (a) x n = Vn(V2 + 3/nÐVl Ð l/n)-+ -f oo as n Ð oo ,since the second multiplier has a positive limit.(c) Xn _ n*(n-VlF+\ ) _ -n* _

1 n+K/iHl1

= Ð n Ð ► Ð oo as n Ð ► oo.

(d) x n = ~ . =

(n 2 Ð n 3 ) 2/3 Ð n y n * Ð n* + n*

1

It means, x n Ð >· 1/3.

(e) Factoring out the terms of the highest power in the numera-tor and denominator, we have:

1.7.5. Find \\mx n if

1/4 > + oo as n Ð ► oo.

n-*

(c) x n =Y l Ð n 3 +n; (d) *¹ = -^ cos n 3 Ð ;

I \ Ð 2n n+l n n(Ð 1)" .

(e) Ð 1 C0S 2« Ð 1 lÐ2n n 2 +l *

(f) x n = j j j- .

1+T+T+-+*

§ 1.8. Testing Sequences for Convergence

Bolzano-Weierstrass* theorem. A monotonic bounded sequence hasa finite limit.


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Theorem on passing to the limit in inequalities. If x n ^y n ^z nand liin x^lim z n = c, then \imy n = c too (c is a number,

n -y cc n ->■ cc n -* cc

-{■ 00 or Ð 00 but not oo).

§ 1.8. Testing Sequences for Convergence

51

1.8.1. Prove that the sequence with the general term x n == (2nÐ l)/(3n+ 1) is an increasing one.

Solution. We have to prove that x n+1 >x n for any n, i.e. toprove that

2/z-j-l 2nÐ l

3/Z + 4 > 3rc+ 1 '

The latter inequality is equivalent to the obvious inequalitye>n 2 + 5n+\ > 6m 2 + 5n Ð 4.

Hence, x n+l > x n .

1.8.2. Given a sequence with the general term

Xn ~ n\ '

Prove that this sequence decreases at n^lO.Solution,

_ 10"+ 1 _ 10" 10 __ 10Xn+1 ~~ (n+ 1)! n\ ' /i+ 1 n+ 1 '

Since < 1 at ft ^10, then x n+1


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\z n \ = \ncosnn\ = n.


Xn Xi Xo Xn i

(a > 1, x > 0) converges.

Solution. Let us prove that this sequence is monotonic andbounded. Firstly, x n < x n _ l as

52


Hence, the given sequence is a decreasing one. Secondly, all itsterms are positive (by condition a > and x > 0), which means

that the sequence is bounded below. Thus, the given sequence ismonotonic and bounded, hence it has a limit.


1 11 i

5+1 1 5 2 +l ^5 3 +l 1 ' · · 1 5»H-1

1,1 1,1,1

^i.e. x 1 Ð b+{ ; x 2 ~ 5+1 + 5 a +1 ; x 2 Ð 5+1+ 5H-1 ^ 5 a +l 'converges.

Solution. The sequence \x n ) increases, since x n+1 = x n + 1/(5" 4-1 + 1)and, hence, #¹ +1 > x n . Besides, it is bounded above, since 1/(5"+ 1) << 1/5" at any n and

1 + _L_ + _^ + +-JÐ <

I *2 i 1 I ^3 l 1 I ' ' ' I Kn _L_ 1 ^

" 5+1 1 5 2 +l 1 5 3 +l 1 ' ' ' 1 5»+l

/ iii ili 1 1/5-1/5" + * _ W, 1\ 1

^ 5 ^ 5 2 ^ 5 3 ^ ' ' ' ^ 5" 1 Ð 1/5 4 V 5" J ^ 4 '

Hence, the sequence converges.


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1.8.6. Taking advantage of the theorem on the existence of alimit of a monotonic bounded sequence, prove that the followingsequences are convergent:

(a) x n = Ðr- ;

(b) * B = 2 + l + l+...+I.

1.8.7. Prove that the following sequences converge and find theirlimits:

(a) x x = V~% * 2 = ^2+l/"2;

n radicals

2 n

(b) *¹

(c) x

(/i + 2)! '

(d) the sequence of successive decimal approximations 1; 1.4;1.41; 1.414; ... of the irrational number j/~2;

(e) x n = n\/n n .

Solution, (a) It is obvious that x x < x 2 < x 3 < . . . < x n <


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find it. Denote

lim x n = y.

n -> oo

Then, x a = V~2

J r x n-v raising to the second power, we obtain

Passing to the limit, we can rewrite this equality as followslim x\ = lim (2 + ,^), or y 2 = 2 + y.

n -*■ oo n -* cc

The roots of the obtained quadratic equation are:

ft = 2; y 2 =Ð 1-

The negative root does not suit here, since x¹>0. Hence, lim x n =y l = 2.

n -* oo

(c) We have ny Ð 1 < E (ny) ^ ny or yÐl^tL^^y. Bui the

sequences j# Ð -^-j and {y} converge, their limit being y, that iswhy lim x n =

(d) This sequence is non-decreasing, since each following termx n+1 is obtained from the preceding one x n by adding one moresignificant digit to the decimal fraction. The sequence is boundedabove, say, by the number 1.5. Hence, the sequence converges, itslimit being V 2.

(e) The sequence decreases monotonically. Indeed,

(fl-f-l)! _ n\ _n\ n n __ n n

Xn+1 ~~ ~~nP ' (n+l)» ~~~ (n+\) n Xn '

Since (n+\)* < l > X » + 1 < X »*

Then, since x n > 0, the sequence is bounded below, hence lim x n

n -+ cc

exists. Let us denote it /. Obviously, /= lim x n ^0. Now let us

n -+ cd

show that / = 0. Indeed,

n n \ n J \ n J n

54



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Hence, ^ n < -y and x n+l


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= 2 M .

On the other hand,l

VrP + n V~n* + n Vri* + n }^n 2 + n

= x ri

Thus,

x n


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lim x n =\ if x n = 2n{Vn 2 +lÐn).

n -* qo


x 1 = Va\ x a =----j/ a + V a\

x s = V0)

has the limit b = {VAa+\ + l)/2.


_ 1 1 l_

* n ~3+l + 3 2 + 2 + ' ' ' ^~?> n + n

has a finite limit.

1.8.15. Prove that a sequence of lengths of perimeters of regular2 w -gons inscribed in a circle tends to a limit (called the length ofcircumference).

§ 1.9. The Limit of a Function

A point a on the real axis is called the limit point of a set Xif any neighbourhood of the point a contains points belonging to Xwhich are different from a (a may be either a proper or an impro-per point).

Let the point a be the limit point of the domain of definitionX of the function f(x). The number A is called the limit of the

function f (x) as xÐ»a, A= lim /(.*:), if for any neighbourhood V

x -* a

of the number A there exists a neighbourhood u of the number asuch that for all x£X lying in a, f(x)£V (the definition of thelimit of a function after Cauchy). The number A may be eitherfinite or infinite. In particular, if the numbers A and a are finitewe obtain the following definition.

56


A number A is called the limit of a function f(x) as xÐ -+a yA = lim f (x), if for any e > there exists a number 8(e) > such


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x -+ a

that for all x satisfying the inequality 0 such that for all x satisfyingthe inequality M(e) and belonging to the domain of definitionof the function f (x) the inequality \f(x) Ð A\ -^2> · · · * · · ·

(belonging to the domain of definition of the function and differingfrom a) the corresponding sequence of values of y

0i = /(*i); 0i = /(*i); 0n = /(*n). ···

has a limit, which is the number A.

1.9.1. Taking advantage of the definition of the limit afterHeine (i.e. in terms of sequences) and of the theorems on the limitsof sequences, prove that

,. 3*+l 1

Solution. Let us consider any sequence x l9 x 2 , ... satisfyingthe following two conditions: (1) the numbers x lt x 2y ... belongto the domain of definition of the function f (x) =(3x+ l)/(5* + 4)(i.e. x n =£ Ð 4/5); (2) the sequence {x n } converges to the number 2,i.e. lim x n = 2.

n -+ 00

To the sequence {x n } there corresponds the sequence of valuesof the function

3^+1 . 3;c a +l .

5*1 + 4 ; 5* 2 + 4 » ' ' '

proceeding from the theorem on the limits (§ 1.7),


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l im fi x )= Hm 3*¹+l_ lim 0^l) a j±i a l¹T.' W ii.5^ + 4- lim (5*¹ + 4) 10 + 4- 2 ·

§ 1.9. The Limit of a Fund ion

57

Thus, independently of the choice of a sequence {x n \ whichconverges to the number 2(x n ^= Ð 4/5), the corresponding sequencesof values of the function f (x n ) converge to the number 1/2, which,according to the definition of the limit of a function, means that

3x+l 15^+4= 2 ·

Note. The definition of the limit after Heine is convenientlyapplied when we have to prove that a function f (x) has no limit.

For this it is sufficient to show that there exist two sequences {x n }and {x n } such that lim x n = lim x'n = a, but the corresponding

n -* qo n -*■ oo

sequences {f(x' n )\ and {f (x¹)\ do not have identical limits.1.9.2. Prove that the following limits do not exist:

1

Solution, (a) Choose two sequences

(a) lim sin 737; C 3 ) ^ m (c) lim sin x.

Y X -+ K Ð*■ CO

1 2

x n = 1 H and x n = 1 + Ta Ð j-tÐ (n = 1 , 2, . . . ),

n 1 nn (4n+l)ji v » » /»

for which

lim x n = lim a:^ = 1.

n -+ n -* oo


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The corresponding sequences of values of the function are:Kx n ) = s\n l + l/{ l nn) _ l =smnn =

and

f «) - sin { + g/[(4w ^ t) n] _ { = sin f£+i ji = sin ( 2/m + f ) = 1 .Hence,

lim / (x n ) = and lim f (x' n ) = 1 ,

1. e. the sequences {/(*¹)} and {/«)} have different limits, whence

it follows that lim sin Ð does not exist.x -* 1 1

(c) Choose two sequences, x n = nn and x' n = 2nn + n/2 (n = 1 ,

2, ...), for which lim jc n = lim a^ = oo. Since

n 00 n -* cc

lim sinx¹= lim sin jt/z = 0,

and

lim sin


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x -* Jt/6

Solution, (a) According to the "e-8" definition we are to provethat for any e > there exists 8 > such that from the inequalityI*Ð 1|such that for all x > M the inequality

will be fulfilled.

Transforming this inequality, we obtain

(e) lim arc tan x = ji/2;

|3a: Ð 8 + 5| = 3|a: Ð 1 | < e.

5* -j-l 5

3x + 9 3

|3* + 9|

14

0, it remains to solve the inequality

§ 1.9. The Limit of a Function


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59

whence

^ 14Ð9e

hence M = 14 7" 98 .

3e

14 g e

Thus, for e > we have found M = Ð ^Ð such that for allvalues of x > M the inequality (*) is fulfilled, and this means that

Let, for example, 8 = 0.01; then M=^ = ^ 1 ^- = 463 -|-.

(c) We have to prove that for any K > there exists 8 >such that from the inequality

|x-l|K.

(i-*) 2

Let us choose an arbitrary number K > and solve the inequality

1

(l-*) 2

whence

K, (**)

|l_x|0).

Thus, if we put 8 = -J=-, then the inequality (**) holds true as


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V K

soon as \x Ð 1 | < 6, which means that lirn n __ . 2 = +oo.

(d) We have to prove that for any K > there exists M >such that from the inequality x > M there always follows the ine-quality \og a x>K. Let us choose an arbitrary number K > andconsider the inequality \og a x> K. If we put a K = M, then atx>M the inequality \og a x>K holds true. Hence,

lim log a *= +oo.

X -»» + 0O

1.9.4. Prove that lim cosjc does not exist.

X -+ 00

1.9.5. Using the sequences of the roots of the equations sin (\/x)=land sin (l/x)= Ð 1, show that the function / (x) = sin (l/x) has nolimit as x Ð >0.

60


1.9.6. Proceeding from Cauchy's definition of the limit of afunction prove that:

(a) lim (3a: Ð 2) = 1; (b) lim x ~ { =2;

K - 1 X - 1 V XÐ 1

(c) lim sin.x: = 0; (d) lim cosx= 1;

x -* o x o

, v ,. 2x Ð 1 2

(e) .'^.s+s'tJ

(f) lim a x = +oo (a > 1);

*-* + 00

(g) lim Ð = 0.

§ 1.10. Calculation of Limits of Functions

I. If the limits \\mu{x) and limy (a;) exist, then the following

x -»■ a x-+a


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theorems hold true:

(1) lim [u (x) ± v (x)} = lim u (x) ± lim v (x);

x -+ a x -* a x -+ a

(2) lim [u (x)-v(x)] = lim u (x) · lim v (x)\

x ^ a x -* a x -* a

lim U (x)

(3) ^ = W^) (Hm»(x)^0).

II. For all main elementary functions at any point of their do-main of definition the equality lim / (x) = / (lim x) = f (a) holds true.

x -+ a x-+a

III. If for all values of a; in a certain neighbourhood of a point a(except for, perhaps, x = a) the functions f(x) and q) (x) are equaland one of them has a limit as x approaches a, then the other onehas the same limit.

IV. The following limits are frequently used:

(1) hm Ð =1;

x-+0 x

(2) lim(l + l/jc)* = lim(l +a) l '* = e = 2. 71828. . .;

x -* a -*

(3) lim lo ^ il+x) = log a e (a>0;a^l);

x^Q x

... ,. In (1 + a:) «

(4) hm / ' = 1;

x^O x

(5) lim ^=^ = lna (a > 0).


61

1.10.1. Find the limits:

i \ i; m ixb + 9x + 7 /u\ r x s + 3x* Ð 9xÐ2


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W I™ 3*. -f*r+T ' (b > I 1 ™ *3_ x _ 6 ··

(C) ^VeZU^' (d) il m ,^_J P and q integers);

(e) Hm ^+^- 3 ; (f)lim^^;

|/*Fp7 Ð 3 J/" 2* Ð 3 . . r f\ x Ð 3

(g) 3/-Xft o 3/0= 5 (h) l0g «

2 r/ x+6 Ð 2 r/ 3xÐ 5 *-3 L

*-+2 j/Ff6-2j/3^5 ' *-*3|_ VHT+6-3J'

> Ð x 2 Ð a:+ 1 r v r + K8a:+ 1

Solution, (a) Since there exist limits of the numerator and deno-

minator and the limit of the denominator is different from zero,we can use the theorem on the limit of a quotient:

lim (4* 5 + 9x+7)lim 4*» + 9* + 7 ^ J ^i V t t ; 4 + 9 + 7 ,

^-13^ + ^+1 lim (3jc 6 + x 3 + 1) 3+1+1

(b) The above theorem cannot be directly used here, since thelimit of the denominator equals zero as xÐ -»2. Here the limit ofthe numerator also equals zero as x Ð »2. Hence, we have the

indeterminate form -jj- . For ^^2 we have

A ;3 + 3x2_9 A ._2 _ (xÐ2) (x 2 + 5x + 1) = jc 2 + 5a:+ 1x 3 Ð xÐ 6 ~~ (xÐ2) (x 2 + 2x + 3) --jc a + 2*+3"

Thus, in any domain which does not contain the point x = 2 thefunctions

r, x £ 3 + 3x 2 Ð 9xÐ 2 , , . x 2 + 5x+l

are equal; hence, their limits are also equal. The limit of thefunction y(x) is found directly:

/vi- * 2 + 5x+ 1 15hm cp(^)^ im = n ;

x -.2 ■ ^2- 2 + 2*+:hence,


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£/ x x 3 + 3x 2 Ð 9xÐ 2 15

X-+2 X-+2 X X O 11

62


(c) Just as in (b), we remove the indeterminate form ^ bytransforming

lim = lim (*+0(^^3-3,) =

-! 3 (1Ð *) ~ i>

1.10.2. Find the limits

X 3 X 2

3a: 2 Ð 4 Ð 3x + 2

(a) lim (

(b) lim (j/9je 2 + l Ð 3x);

* -* + 00

(c) lim 2^+3^/7+5^/7

v^ + oo V 3*Ð 2+ 3 /2xÐ 3

(d) lim 0/2x 2 Ð 3Ð 5x);

* -* Ð 00

(e) lim x(Vx 2 +l Ð x)\

X -* + QO

/f , ,. V2x^f3 , ]/~2x 2 +3

(g) lim S^** 3 *.

X -* OO

So/««on. (a) lim (^-^) .


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Here we have the indeterminate form oo Ð oo; let us subtract thefractions

lim f^_^)=lim 2x3+4x2 -

jc^oo \3* 2 Ð 4 3x + 2 y ^ 9x 3 + 6a: 2 Ð \2xÐ 8

= lim 2 + 4/x 2

^.Too 9 + 6/a:Ð 12/a: 2 Ð 8/a: 3 9 '

Note. We see that in such examples the limit is equal to theratio of the coefficients at the superior power of x (provided thepolynomials are of the same degree).

/ux ,. (}^9x 2 +i Ð 3x) ,. 1 n

(b) lim Ð -J- ' = lim r = 0.

+ 1 x -+ + cc J^-j- 1+3*

(c) In handling such examples bear in mind that the function

f{x)= \/p n (x), where p n (x) is a polynomial ofjdegree n, tendingto infinity in the same way as the function "j/ x n . This allows usto single out the superior power of x and divide both the nume-rator and denominator by this power of x. In the given example

§ 1. 10. Calculation of Limits of Functions

63

the divisor is [/ x\ then we obtain:

2J/T+3 j/T _ 2 + 3/*/T +5/ff^

+ j/3xÐ 2 + l/2xÐ3 x^ + oo |/"3 Ð 2/x+ £/4/xÐ 12/x 2 + 9/x 3

2

(d) Since the sum of two positive infinitely large quantities isalso an infinitely large quantity, then

Urn (1/2j?^3Ð 5*)= lim [V2x* Ð 3 + (Ð5*)] -

X -> - cc X-*-oo

(f) At a; > we have |/# 2 = .x;, therefore


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, . ^*«(2 + 3/*«) _ *V~2 + 37F _

+ "","™ ao *(4 + 2/*) ~ 4 '

At a: < we have \^x 2 = Ð x and, hence,

, im y* a (2 + 3/**) = Hm ^2 + 3/^ _ ]/"2~

*(4 + 2/x) *(4 + 2/*)

/Vote. From this it follows, incidentally, that lim ^ * x \ ~t 3 doesnot exist.

lim 2x/(* + 3)

(g) lim5 2 ^* +3 ) = 5*- fl0 = 5 2 = 25.

* -*· oo


(a) Hm^T&l ; (b) H ra _*±I ;

(c) hm < ,^5 X - ; (d) lim-*- Ð (k positive in-

x-+0

teger);

, x ,. sin (xÐ ji/6)

^3 Ð 2cosx' U *-jt/2 sin*) 2 '

, v y 2 sin 2 x-f sin xÐ 1*i™/6 2sin 2 xÐ 3sinx+l '

Solution (method of substitution), (a) Let us put 26 + x = z 3 .Then x=^z*Ð 26 and 2Ð^3 as * 1; hence

lim 3/ ^Z_ 2 = lim ^ - lim ^-3)(^+3*+9) _

= lim2(z 2 + 3z + 9) = 54.

2^3

64


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(d) Let us put l+x=z k ; then x=z k Ð 1 and 2Ð^1 as x Ð + 0.Hence,

^ T ^~ l -i; m _£z± * ( S ee Problem 1.10.1 (d)).

lim

X-y

2 -* 1

(e) Let us put x Ð ji/6--=2; then x = z + n/S and z Ð ► () asx Ð On substituting we obtain

lim y^U limÐ sin2

lim

t-jT/6 l/"3 Ð 2 cos a: 2-0 /"3 Ð 2 cos (z -f ji/6)sin z 2 sin (z/2) cos (2/2)

= lim

= lim A _ CQS(2/2) =1.

z - o y 3 sin (2/2) + cos (z/2)

>-o J^~3 Ð "K" 3 cos 2+ sin 2 2 - o 2 /" 3 sin 2 (z/2) + 2 sin (z/2) cos (z/2)


, x ,. 1 Ð cos x /L v i · tan x Ð sin a:(a) hm Ð -5 Ð ; (b) lim

x^O x

(c) lim cos^2)


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x-0

X 3

a:- 1

f /vi- 1 Ð cos a: 2 sin 2 (x/2)

Solution, (a) lim 5 Ð = lim r^'-

*-»o x *-»o x

/ux ,. tan a: Ð sinA; sin a: (1 Ðcos a:)

(b) lim - = lim ' Ð

l Hm / sin(^/2) y ==1 ,2 J™ ^ x/2 J 2 '

*-*

C0S x ' x *

= lim-

1 sin* 1 Ð cos a; 1

cos x

x 2

(c) Let us put 1 Ð x = z. Then x=l Ð z and z Ð ►O as x Ð ► 1.Hence,


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COS -y X

lim - Ð Ð = lim

/ Jl TC \

1Ð X

= lim

sin -7- z

2 3T

2-0

2-

Note. For a simpler method of solving similar problems see § 1.12.1.10.5. Find the limits:

(a) lim(l + l/*) 7 *;

(c ^ ra .fe)* ;

(b) lim(l +x) l ^\

x-»

(d) lim(l+£/A:) m *;

(e) lim

In (\+x) t3*Ð 1 1


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In (a+x) Ð In a t

x-+ o(g)

x ->

... In x Ð 1

(0 llm -r=r ·

(f) lim

*-*

(h) lim

e 4 *Ð 1 .tan a: *


65

Solution, (a) lim H Ð y x = Hni

1 +

1 \*

lim 1 +

= e'\


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(e) lim = lim

In (1 +x)

3* Ð 1

1

ln3'

x -* u " 1 x -v

(i) Put jt/eÐ l=z; then x--=e(z+l); zÐ *0 as xÐ*e. On substi-

tuting we obtain

Hm lni = J = lim ln(x/ e)

,,-ilin, 1^1+^1Ð 1) e z _ z e

1.10.6. Find

·im (1+^

Solution, lim (l+l) x =lim |^1 +1^*] = 1.


(a) lim/l-MV 1 -^ 1 -^.

(b) lim

x 2J r 2 X Ð 1 \(2* + 1 >/( atÐ 1)


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2x 2 Ð 3a: Ð V

Solution, (a) Denote:

fW = (l+x)/(2 + x);

/ \ 1Ð ^*

Hm^(x) = Hml±^ = -|;lim (p (x) = lim ^ ^~ x = ir .

x-*\ x->\ l ~ x 1

But at finite limits lim / (x) = A > 0, lim q) (.*:) = 5 the followingrelation holds true:

lim (p (x) In / (x)

lim [/ {x) = e x ~+ a =e B\nA == j[B t

1

Hence,

lim

°- r ""'"-"-(!)" ! =/!

/Vo/e. If in handling examples of the form lim [/ (x)]v {x) it turns

x-+a

out that Vunf(x)=l and lim cp (jc) = oo, then the following

3Ð 31 43

66


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transformation may be recommended:

lim [f(x)]* = lim {1 + [/ (*)Ð 1 }}* {x) -

x^-a x -a

l«»n(p(x)L/U)-l]

= lim { [ 1 + (/ (x) Ð 1 )] - 1 >}* U - n = e**a . (*)


(a) lim( 2 ^r +3 ; (b) li.n(4±^) ,/,m *;

(c) lim (1 +sin :rix) cot ™;

(d) lim (f|^~y /U Q) (a¥=kn> with & an integer).Solution, (a) Let us denote:

^^StT' ^ (·*>-!]

/WÐ I=|S±4 Ð 1 :

lira q> (x) [/(*)-!] = - lim 2 gf + 3) = - 8.

Therefore

1; /2xH-3\8x« + 3

1.10.9. The function /(#) is given with the aid of the limit


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/? -> cc x r 1

Investigate this function and graph it.Solution. Consider three cases:

(1) \x\> 1. Since in this case lim x 2n --= 00, then

(2) |jc| 1

f(x)=\-l if \x\


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have increasedl

\ 50

times. Taking into account that

1

e.

lim Ð J =e> we can approximately consider that

Hence, after 100 years the population of the country will have

increased e' 1 « 7.39 times.Of course, this estimation is very approximate, but it gives an

idea as to the order of the increase in the population; ^the quan-

/ 1 \ioo \

tity ^1 =7.245 to within three decimal places).

50/


, x i- cos x-\~A tan x(a) lim- 1

2 Ð x Ð 2x*

(b) z /__ x _ 2 i

(c) lirn^E^;

2x- Ð 5x + 4 .5a- 2 Ð 2x Ð 3 *

(d) lim

68


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(e) lini(j/>+ 1Ð Ð 0;

X - > QO


(c) lim . s '"., a , ; (d) lim tan 2x tan (n/4 Ð x);

gc-»ji 1 ~~ a "' n *-*ji/4. .. tan 3 x Ð 3 tan x(e) ™ 3 cos (x+n/6) *


(a) lim (1 + 4/x)* +3 ; (b) lim *"* -1 ;

(c) lim^=^; (d) lim (1 + 3 tan 2 x) cot2 *;

x->0 * *->0

(e) lim (sin2*)t- 22 *; (f) lim lg-r\) ;

x-yn/4 x-y cd \ zx -r 1 /

(g) lim (tan ;c) tan 2 *; (h) lim (sin x) ian x \

X-+JI/2 X-yJt/2

(k) hm Ð .

x +0 x


. , ,- arc cos (1Ð^) In tan a:

(a) i nn ; (b) hm ;

(c) lim Ð Ð In (1 + a sin a:).

o sin x

§ I. II. Infinitesimal and Infinite Functions.

Their Definition and Comparison

The function a(x) is called infinitesimal as or as .v-^ oo

if lim a (x) = or lima (*) = 0.

x-*a X-y CD

The function f (x) is called infinite as ^->a or as x-oo iflim/(A;) = oo or lim / (a;) = oo.


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§ 1.11. Infinitesimal and Infinite Functions

69

A quantity inverse to an infinite quantity is called an infinite-simal.

Infinitesimal functions possess the following properties:

(1) The sum and the product of any definite number of infinite-simal functions as x a are also infinitesimals as x -> a.

(2) The product of an infinitesimal function by a bounded functionis an infinitesimal.

Comparison of Infinitesimals. Let the functions a(x) and P (x) beinfinitesimal as x a. If

, . a (x)lim q-K- =*-*a P to

where c is a certain finite number different from zero, then thefunctions a (x) and P (x) are called infinitesimals of the same order.If c=l, then the functions a(x) and P (x) are called equivalent;notation: a(x)~$(x).

If £ = 0, then the function a(x) is called an infinitesimal of a/ug/ier order relative to P (*), which is written thus: a (x) = o (P (x)),and P(jc) is called an infinitesimal of a /oayer order with respect

to a (x).

If lim R a /f where 0

(b) f(x) = (x Ð l) 2 sin 3 as a: 1 are infinitesimals.Solution, (a) It is sufficient to find the limit


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lim / (x) = lim 2x ~\ = 0.* ► 2 « -> 2 r o

(b) Firstly, the function cp(x) = (x Ð l) 2 is infinitesimal as x 1;indeed, lim (a: Ð l) 2 Ð 0. Secondly, the function

X > 1

\J)(a:) - sin 3 Ð j ; 1,

is bounded:

I sin 3 Ð < 1.

Hence, the given function f (x) represents the product of thebounded function \p(x) by the infinitesimal (p(x), which means thatf(x) is an infinitesimal function as x -+ 1.

70

Ch. 1. Introduction to Mathematical Analysis

1.11.2. Prove that the functions

(a) /(*)= ^-yi as x -+ 4 >

(b) f (x) = as x Ð ooare infinitesimal.

1.11.3. Find

lim x sin (l/x).x - o

Solution. Since x is an infinitesimal as x Ð ► and the function

sin (l/x) is bounded, the product xs\n(\/x) is an infinitesimal, which

means that lim x sin (l/x) = 0.* -* o

1.11.4. Compare the following infinitesimal functions (asxÐ ► ())with the infinitesimal (p(x) = A;:

(a) /, (x) = tan x 3 ; (b) / 2 (x) = ;

(c) U*) = K9 + x-3.Solution, (a) We have


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l im ifE^ = li m

tan a: 3 ,1 ,. tan* 3

■x 2 l = li

J X -

lim Ð ^ Ð liin a: 2 = 0.

x -> * x -

Hence, tanx 3 is an infinitesimal of a higher order relative to x.

(b) We have

* - * x -* J/ X 2 3/^

Hence, jj/sin 2 * is an infinitesimal of a lower order as comparedwith A'.

(c) We have

lim Ð Ð Ð = li in r = ir.

x-> o x x-+o + 6

Hence, the infinitesimals V~9-{-x Ð 3 and x are of the same order.

1.11.5. Determine the order of smallness of the quantity p withrespect to the infinitesimal a.

(a) p=rcosa Ð cos2a; (b) P = tana Ð sina.

Solution, (a) p^=cosa Ð cos 2a = 2 sin -| a sin

Whence

|in] J^ =lim 2sin(3a/2)sin ( a/2) = 3_

a o « 2 a - o « 2 2

Hence, P is an infinitesimal of the same order as a 2 , i. e. of thesecond one with respect to a.

§ 1.12. Equivalent Infinitesimals

71


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1.11.6. Assuming * Ð ► oo, compare the following infinitely largequantities:

(a) / (*) = 3* 2 + 2* + 5 and cp (x) = 2* 3 + 2x Ð 1 ;

(b) f(x) = 2x 2 + 3x and cp (*) = (* + 2) 2 ;

(c) / (x) = Y x + a an d oo 2* 3 + 2xÐ 1 2 + 2/x 2 Ð 1/x 3

1.11.7. Prove that the infinitesimalsa=* and P=*cos(l/*)(as*Ð ^0)are not comparable, i. e. their ratio has no limit.

X cos ( 1 1 x\

Solution. Indeed, lim y -L-L= lim cos (l/x) does not exist (prove

x -> o * x -> o

it!), which means that these infinitesimal functions are not com-parable.

1.11.8. If * Ð ► 0, then which of the following infinitesimals is(are) of a higher order than x\ of a lower order than x; of thesame order as *?

(a) 100*; (b) * 2 ; (c) 6 sin*; (d) sin 3 *; (e) j/tan 3 *.

1.11.9. Let x Ð ► O. Determine the orders of the following infini-tesimal functions with respect to x:

(a) 2 sin 4 a;Ð* 5 ; (b) [/sin 2 * + * 4 ;

(c) Vl + x 3 Ð 1; (d) sin 2* Ð 2 sin*;

(e) 1 Ð 2cos(x+£); (f) 2 |/sin*;

(g) TZTf; ( h ) tan* + * 2 ;

(i) cos* Ð l/cosx; (j) Ð cos*.

1.11.10. Assuming the side of a cube to be an infinitesimal, de-termine the order of smallness of the diagonal of the cube (d), of

the area of its surface (S); of its volume (V).

§ 1.12. Equivalent Infinitesimals.Application to Finding Limits

If the functions a (*) and P (*) are infinitesimal as *Ð and ifa ( x ) ~ y ( x )> P (*) ~ 6 (*), then

lim -77t4-= Hm -^y- (replacing an infinitesimal by an equivalent one).


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72


If

lim f(x) = k f < |*| < oo,

x -* a

then

/ (x) a (x) ~ ka (x).

If

a.(x)~y(x) 9P(*)~Y(*),

then

a (x) ~ P (x).

For two infinitesimal functions to be equivalent it is necessaryand sufficient that their difference be an infinitesimal of a higherorder as compared with each of the two.

Listed below are infinitesimal functions:

(a(x) is an infinitesimal as xÐ +0)

(1) sin a (x) ~ a (x); (2) tan a (x) ~ a (x);

(3) 1Ð cosa(x) ~ [a(Jc)] 2 /2;

(4) arc sin a (x) ~ a (x)\ (5) arc tan a (x) ~ a (x);(6) ln[l+a(x)] ~a(x); (7) a a{x} Ð 1 ~ a (x) \na(a > 0), in particular, e* U) Ð 1 ~ a (x);

(8) [l+a(x)] p Ð 1 - Pa(x), in particular, (V l+aW-l- a ^.1.12.1. Prove that as x-^0

< a > l ~YTTi~'* x; (b) 1_ TT7~* ;(c) sin

Solution, (a) By formula (8) at P=l/2 we have

1 1 o/T+^-D^i.

X


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X.

(c) By formula (1) we have

sin]/ xV~x '~y ' x)/"x ' = x 3


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/ \ r In cos a: ,,v 1; V^" 1 -f- a: -f- a: 2 Ð 1 ,

(c) hm 4 Ð ; (d) hm \ ^ ;

, x ,. sin 2*4- arc sin 2 *Ð arc tan 2 *

(e) hm ±- 3- ;

.pv |. 3 sin* Ð * 2 + * 3]" n tan* + 2sin 2 * + 5* 4 ;

(g) Hm ^ sm x ~ tan ^ 2 ~^ 1 ~~ cos 2 x )*-r xb .

x ^ o 7 tan 7 x-\- sin 6 x-j-2 sin 5 xsin j/ X * In (1 +3*)

(h) lim Ð 3 , ,

^°(arctan |/* )■ (* 5 / * - 1 )

j. 1 Ð cos *+2sin xÐ sin 3 xÐ * 2 + 3* 4' ' x ^ tan 3 * Ð 6 sin- x -}- * Ð 5* 3

Solution, (a) We have sin5A:^5x; In (^1 + 4x) ■ Ð ■ 4x (see the listof equivalent infinitesimals on page 72). Therefore

sin 5* ,. 5* 5

J™lF(T+W = , Il i ,1 o-4F = T-

(c) lim *2»JL- = lim ln " + S*-' )1 =

¹ cos* Ð I A ,. x 2 /2

= 4 hm 7i = Ð 4 hm Ð Ð = Ð2.

a: -> * x -> o x

(d) From the list of equivalent infinitesimals we find:

V 1 +x + x 2 Ð 1 - (x -\- x 2 )/2 ~ x/2, sin \x ~ \x.

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Ch. I. introduction to Mathematical Analysis

Therefore


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vT+7+x 5 Ð l r x/2 I

lllTl rÐ -j Ð lUTl Ð Ð = Ð

(e) Using the list of equivalent infinitesimal functions given onpage 72 we obtain

sin 2x + arc sin 2 x Ð arc tan 2 x ~ sin 2x ~ 2x.

Hence,

,. sin 2x-\- arc sin 2 x Ð arc tan 2 x r 2x 2

hm Yx = Ilrn 35 = t ·

/7~ Y~x\ In (1 + 3x) - 3x;

arc tan K7; e 5 ^* Ð 1 Ð 5 J^jc;

lim sin^ln(l+3,_) m ^.3,_3

' (arc tan V~xY

1.12.4. Find the approximate values of the roots (/ 1.02 and1^0.994. Estimate the absolute error.

Solution. Use the approximate formula

]/"l+x ~l+x/2 (*)(for x sufficiently close to zero). In our case

J/1 + 0.02- 1+^= 1.01;

V 1Ð0.006 ~ 1 Ð = 0.997.

To estimate the error we note that

|_(Kr+7- 1) =■! (*- 2 ^1+^+ 2) =

= j(*+i- 2 K^+i) = l(^^

Hence, the absolute error of the approximate formula (*) is esti-

X 2

mated by the quantity y.

Using this estimate we find that the absolute error of the root1^1762 « 1.01 is = 0.00005, and the absolute error of

j/0994 ^ 0.997 amounts to « ( -^^ « 0.000005.

o

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