1.Alternate Edition Calculuswith analytk: geometry Earl W Swokowski Marquette University ~ Prindle, Weber &Schmidt Boston, Massachusetts

2. Dedicated to the memory ofmy mother and father; Sophia andjohn Swokowski F='WS F='UBLISHEF=tS Pnndle Weber & Schmrdt II.' Wollard Granl Press ooc: Duxbury Press Statler Off1ce Buld.ng 20 Provdence Street Boston Massachusetts 02116 Copyright 1983 by PWS Publishers All rights reserved. No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system, without permission, in writing, from the publisher. PWS Publishers is a division of Wadsworth, Inc. Portions of this book previously appeared in Calculus with Ana- lytic Geometry, Second Edition by Earl W. Swokowski. Copyright 1979 by Prindle, Weber & Schmidt. 87 86 85 84 83 - 10 9 8 7 6 5 4 3 2 ISBN 0-87150-341-7 Library of Congress Cataloging in Publication Data Swokowski, Earl W. Calculus with analytic geometry. Includes index. I. Calculus. 2. QA303.S94 1983 ISBN 0-87150-341-7 Geometry, Analytic. 515'.15 82-21481 I. Title Cover image courtesy of General Motors Research Laboratories. The computer graphic image depicts the location of valence elec- trons trapped near the surface of rhodium. Quantum mechanical calculations using the Schrodingerequation were employed to gen- erate the image. The two sets of peaks in the foreground reveal a preferential accumulation of electrons around the surface atoms. Production and design: Kathi Townes Text composition: Composition House Limited Technical artwork: Vantage Art, Inc. Cover printing: Federated Lithographers-Printers, Inc. Text printing/binding: Von Hoffmann Press, Inc. Printed in the United States of America 3. Preface Most students study calculus for its use as a tool in areas other than mathematics. They desire information about why calculus is important, and where andhow it can be applied. I kept these facts in mind as I wrote this text. In particular, when introducing new concepts I often refer to problems that are familiar to students and that require methods of calculus for solutions. Numerous examples and exercises have been designed to further motivate student interest, not only in the mathematical or physical sciences, but in other disciplines as well. Figures are frequently used to bridge the gap between the statement of a problem and its solution. In addition to achieving a good balance between theory -and applications, my primary objective was to write a book that can be read and understood by college freshmen. In each section I have striven for accuracy and clarity of exposition, together with a presentation that makes the transition from precalculus mathematics to calculus as smooth as possible. The comments that follow highlight some of the features of this text. A review of the trigonometric functions is contained in the last section of Chapter I. It was placed there, instead of in an appendix, to alert students to the fact that trigonometry is, indeed, a prerequisite for calculus, as indicated by the title of the chapter. Tests for symmetry are also introduced early, so that they can be used throughout the text. In Chapter 2 limits involving the sine and cosine func- tions are considered after limits of algebraic functions, and thus are readily available for use in obtaining derivative for- mulas in Chapter 3. The early introduction of trigonometric functions leads to some nontrivial applications of the Chain Rule and enlarges the scope ofapplications ofthe derivative. In Chapter4testvalues are used to determine intervals in which derivatives are positive or negative. This pedagogical device is also employed to help obtain graphs of rational functions. Chapters 5 and 6, on properties and applications ofdefi- nite integrals, include exercises on numerical integration that require reference to graphs to approximate areas, vol- umes, work, and force exerted by a liquid. Inverse functions are discussed in the first section of Chapter 7 and are used in Section 7. 3 to define the natural exponential function as the inverse of the natural log- arithmic function. Chapters 8- 10 contain material on transcendental func- tions, techniques of integration, and improper integrals. Infinite series are presented in a precise manner in Chapter II. Chapter 12 consists of a detailed study of conic sections. Chapters 13-15 deal with curves, vectors and vector- valued functions. There are many examples and exercises pertaining to parametric and polar equations, and a strong emphasis is placed on geometric and applied aspects of vectors. Functions of several variables are discussed at length in Chapter 16. The relevance of level curves and surfaces to practical situations is illustrated in examples and exercises. The approach to increments and differentials is motivated by analogous single variable concepts. The definition of direc- tional derivative does not require the use of direction angles ;;; 4. iv Preface of a line, and considerable stress is given to the gradient of a function. The study of maxima and minima includes an examination of boundary extrema. The final section, on Lagrange multipliers, includes a proof that indicates the geometric nature of why the method is valid. Properties and applications of multiple integrals are con- sidered in Chapter 17. Vector fields are discussed in Chapter 18, and special attention is given to conservative fields. The physical sig- nificance of divergence and curl is brought out by using the theorems of Gauss and Stokes. The last two sections con- tain results on Jacobians and change of variables in multiple integrals. Chapter 19, on differential equations, includes two sepa- rate sections on applications. There is a review section at the end of each chapter con- sisting of a list of important topics and pertinent exercises. The review exercises are similar to those that appear throughout the text and may be used by students to prepare for examinations. Answers to odd-numbered exercises are given at the end ofthe text. Instructors may obtain an answer bookletfor the even-numbered exercises from the publisher. Portions of this text are based on material that appears in my book Calculus withAnalyticGeometry, Second Edition. This second edition is available for courses where a later introduction of the trigonometric functions is desired. I wish to thank the following individuals, who received all, or parts of, the manuscript and offered many helpful suggestions: Alfred Andrew, Georgia Institute of Technol- ogy; Jan F. Andrus, University of New Orleans; Robert M. Brooks, University of Utah; Dennis R. Dunniger, Michigan State University; Daniel Drucker, Wayne State University; Joseph M. Egar, Cleveland State University; Ronald D. Ferguson, San Antonio State College; Stuart Goldenberg, California Polytechnic State University; Theodore Guinn, University of New Mexico; Joe A. Guthrie, University of Texas, El Paso; David Hoff, Indiana University; Adam Hulin, University ofNew Orleans; W. D. Lichtenstein, Uni- versity of Georgia; Stanley M. Lukawecki, Clemson Uni- versity; Louise E. Moser, California State University, Hayward; Norman K. Nystrom, American River College; David A. Petrie, Cypress College; William Robinson, Ven- tura College; JohnT. Scheick, OhioState University; Jon W. Scott, Montgomery College; Monty J. Strauss, Texas Tech University; Richard G. Vinson, University of South Ala- bama; Loyd Wilcox, Golden West College; and T. J. Worosz, Metropolitan State College, Denver. I also wish to express my gratitude to Christian C. Braunschweiger of Marquette University, who provided an- swers for exercises; Thomas A. Bronikowski of Marquette University, who authored the student supplement containing detailed solutions for one-third of the exercises; Stephen B. Rodi ofAustin Community College, who developed a com- plete solutions manual; Michael B. Gregory of the Univer- sity ofNorth Dakota, who supplied a number of challenging exercises; and ChristopherL. Morgan, California State Uni- versity at Hayward, and Howard Pyron, University of Mis- souri at Rolla, who prepared the computer graphics. Special thanks are due to Stephen J. Merrill ofMarquette University for suggesting several interesting examples, including one that indicates how infinite sequences and series may be em- ployed to study the time course of an epidemic, and another that illustrates the use ofexponential functions in the field of radiation therapy. I am grateful for the valuable assistance of the staff of PWS Publishers. In particular, Mary LeQuesne and Joe Power were very helpful with exercise sets; Kathi Townes did a superlative job as copy editor; and David Pallai, who supervised the production of this large project, was a con- stant source of information and advice. In addition to all of the persons named here, I express my sincere appreciation to the many unnamed students and teachers who have helped shape my views on how calculus should be presented in the classroom. Earl W. Swokowski 5. Table ofContents Introduction: What Is Calculus? ix 1 Prerequisites for Calculus 1 1.1 Real Numbers 1 1.2 Coordinate Systems in Two Dimensions 9 1.3 Lines 18 1.4 Functions 24 1.5 Combinations of Functions 33 1.6 The Trigonometric Functions 37 1.7 Review 47 2 Limits and Continuity ofFunctions 49 2.1 Introduction 49 2.2 Definition of Limit 54 2.3 Theorems on Limits 60 2.4 One-Sided Limits 68 2.5 Limits of Trigonometric Functions 72 2.6 Continuous Functions 76 2.7 Review 86 3 The Derivative 87 3.I Introduction 87 3.2 The Derivative of a Function 92 3.3 Rules for Finding Derivatives 98 3.4 Derivatives of the Sine and Cosine Functions 106 3.5 Increments and Differentials 111 3.6 The Chain Rule 119 3.7 Implicit Differentiation 126 3.8 Derivatives Involving Powers of Functions 131 3.9 Higher Order Derivatives 135 3.10 Newton'sMethod 138 3.11 Review 141 4 Applications ofthe Derivative 144 4.1 Local Extrema of Functions 144 4.2 Rolle's Theorem and the Mean Value Theorem /52 4.3 The First Derivative Test 156 4.4 Concavity and the Second Derivative Test 162 4.5 Horizontal and Vertical Asymptotes 170 4.6 Applications of Extrema 182 4.7 The Derivative as a Rate of Change 192 4.8 Related Rates 201 4.9 Antiderivatives 206 4.10 Applications to Economics 214 4.11 Review 22/ 5 The Definite Integral 223 5.1 Area 223 5.2 Definition of Definite Integral 232 5.3 Properties of the Definite Integral 239 5.4 The Mean Value Theorem for Definite Integrals 244 5.5 The Fundamental Theorem of Calculus 246 5.6 Indefinite Integrals and Change of Variables 254 5.7 Numerical Integration 263 5.8 Review 271 v 6. vi Table ofContents 6 Applications ofthe Definite Integral 274 6.1 Area 274 6.2 Solids of Revolution 283 6.3 Volumes Using Cylindrical Shells 291 6.4 Volumes by Slicing 296 6.5 Work 299 6.6 Force Exerted by a Liquid 306 6.7 Arc Length 31/ 6.8 Other Applications 317 6.9 Review 323 7 Exponential and Logarithmic Functions 325 7.1 Inverse Functions 325 7.2 The Natural Logarithmic Function 329 7.3 The Natural Exponential Function 337 7.4 Differentiation and Integration 345 7.5 General Exponential and Logarithmic Functions 352 7.6 Laws of Growth and Decay 359 7.7 Derivatives of Inverse Functions 366 7.8 Review 370 8 Other Transcendental Functions 372 8.1 Derivatives of the Trigonometric Functions 372 8.2 Integrals of Trigonometric Functions 378 8.3 The Inverse Trigonometric Functions 382 8.4 Derivatives and Integrals Involving Inverse Trigonometric Functions 387 8.5 The Hyperbolic Functions 393 8.6 The Inverse Hyperbolic Functions 399 8.7 Review 403 9 Additional Techniques and Applications ofIntegration 405 9.1 Integration by Parts 406 9.2 Trigonometric Integrals 412 9.3 Trigonometric Substitutions 418 9.4 Partial Fractions 423 9.5 Quadratic Expressions 430 9.6 Miscellaneous Substitutions 433 9.7 Tables of Integrals 437 9.8 Moments and Centroids of Plane Regions 440 9.9 Centroids of Solids of Revolution 447 9.10 Review 453 10 Indeterminate Forms, Improper Integrals, and Taylor's Formula 457 10.1 The Indeterminate Forms 0/0 and xfx 457 10.2 Other Indeterminate Forms 464 I0.3 Integrals with Infinite Limits of Integration 468 10.4 Integrals with Discontinuous Integrands 473 10.5 Taylor's Formula 479 10.6 Review 488 11 Infinite Series 490 11.1 Infinite Sequences 490 11.2 Convergent or Divergent Infinite Series 500 11.3 Positive Term Series 5// 11.4 Alternating Series 519 11.5 Absolute Convergence 523 11.6 Power Series 530 II. 7 Power Series Representations of Functions 536 11.8 Taylor and Maclaurin Series 541 11.9 The Binomial Series 550 11.10 Review 553 12 Topics in Analytic Geometry 556 12.1 Conic Sections 556 12.2 Parabolas 557 12.3 Ellipses 565 12.4 Hyperbolas 571 12.5 Rotation of Axes 577 12.6 Review 581 13 Plane Curves and Polar Coordinates 583 13. I Plane Curves 583 13.2 Tangent Lines to Curves 59/ 13.3 Polar Coordinate Systems 594 13.4 Polar Equations of Conics 603 13.5 Areas in Polar Coordinates 608 13.6 Lengths of Curves 611 13.7 Surfaces of Revolution 615 13.8 Review 6/9 14 Vectors and Solid Analytic Geometry 621 14.1 Vectors in Two Dimensions 621 14.2 Rectangular Coordinate Systems in Three Dimensions 631 14.3 Vectors in Three Dimensions 635 14.4 The Vector Product 645 7. 14.5 Lines in Space 652 14.6 Planes 654 14.7 Cylinders and Surfaces of Revolution 661 14.8 Quadric Surfaces 665 14.9 Cylindrical and Spherical Coordinate Systems 670 14.10 Review 673 15 Vector-Valued Functions 676 15.1 Definitions and Graphs 676 15.2 Limits, Derivatives, and Integrals 680 15.3 Motion 688 15.4 Curvature 692 15.5 Tangential and Normal Components of Acceleration 700 15.6 Kepler's Laws 705 15.7 Review 710 16 Partial Differentiation 713 16.1 Functions of Several Variables 713 16.2 Limits and Continuity 72/ 16.3 Partial Derivatives 727 16.4 Increments and Differentials 733 16.5 The Chain Rule 742 16.6 Directional Derivatives 750 16.7 Tangent Planes and Normal Lines to Surfaces 758 16.8 Extrema of Functions of Several Variables 764 16.9 Lagrange Multipliers 770 16.10 Review 778 17 Multiple Integrals 780 17. I Double Integrals 780 17.2 Evaluation of Double Integrals 785 17.3 AreasandVolumes 794 17.4 Moments and Center of Mass 798 17.5 Double Integrals in Polar Coordinates 804 17.6 Triple Integrals 809 17.7 Applications of Triple Integrals 816 17.8 Triple Integrals in Cylindrical and Spherical Coordinates 820 17.9 Surface Area 824 17.10 Review 827 Table of Contents vii 18 Topics In Vector Calculus 829 18.1 Vector Fields 829 18.2 Line Integrals 836 18.3 Independence of Path 847 18.4 Green's Theorem 854 18.5 Surface Integrals 862 18.6 The Divergence Theorem 869 18.7 Stokes' Theorem 875 18.8 Transformations of Coordinates 882 18.9 Change of Variables in Multiple Integrals 885 18.10 Review 891 19 Differential Equations 893 19. I Introduction 893 19.2 Exact Differential Equations 898 19.3 Homogeneous Differential Equations 902 19.4 First-Order Linear Differential Equations 906 19.5 Applications 909 19.6 Second-Order Linear Differential Equations 914 19.7 Nonhomogeneous Linear Differential Equations 920 19.8 Vibrations 926 19.9 Series Solutions of Differential EquatiDns 931 19.10 Review 934 Appendices Mathematical Induction Al II Theorems on Limits and Definite Integrals A8 III Tables A Trigonometric Functions A18 B Exponential Functions Al9 C Natural Logarithms Al9 IV Formulas from Geometry A20 Answers to Odd-Numbered Exercises A21 Index A57 8. Introduction: What is Calculus? Calculus was invented in the seventeenth century to provide a tool for solving problems involving motion. The subject matter of geometry, algebra, and trigonometry is applicable to objects which move at constant speeds; how- ever, methods introduced in calculus are required to study the orbits of planets, to calculate the flight of a rocket, to predict the path of a charged particle through an electromagnetic field and, for that matter, to deal with all aspects of motion. In order to discuss objects in motion it is essential first to define what is meant by velocity and acceleration. Roughly speaking, the velocity of an object is a measure of the rate at which the distance traveled changes with respect to time. Acceleration is a measure of the rate at which velocity changes. Velocity may vary considerably, as is evident from the motion of a drag-strip racer or the descent of a space capsule as it reenters the Earth's atmosphere. In order to give precise meanings to the notions of velocity and acceleration it is necessary to use one ofthe fundamental concepts ofcalculus, the derivative. Although calculus was introduced to help solve problems in physics, it has been applied to many different fields. One of the reasons for its versatility is the fact that the derivative is useful in the study of rates of change of many entities other than objects in motion. For example, a chemist may use derivatives to forecast the outcome of various chemical reactions. A biologist may employ it in the investigation of the rate of growth of bacteria in a culture. An electrical engineer uses the derivative to describe the change in current in an electrical circuit. Economists have applied it to problems involving corporate profits and losses. The derivative is also used to find tangent lines to curves. Although this has some independent geometric interest, the significance of tangent lines is of major importance in physical problems. For example, if a particle moves along a curve, then the tangent line indicates the direction of motion. If we restrict our attention to a sufficiently small portion of the curve, then in a ix 9. X Introduction: What is Calculus? certain sense the tangent line may be used to approximate the position of the particle. Many problems involving maximum and minimum values may be attacked with the aid of the derivative. Some typical questions that can be answered are: At what angle of elevation should a projectile be fired in order to achieve its maximum range? If a tin can is to hold one gallon of a liquid, what dimensions require the least amount of tin? At what point between two light sources will the illumination be greatest? How can certain corporations maximize their revenue? How can a manufacturer minimize the cost of producing a given article? Another fundamental concept of calculus is known as the definite integral. It, too, has many applications in the sciences. A physicist uses it to find the work required to stretch or compress a spring. An engineer may use it to find the center of mass or moment of inertia of a solid. The definite integral can be used by a biologist to calculate the flow of blood through an arteriole. An economist may employ it to estimate depreciation ofequipment in a manufacturing plant. Mathematicians use definite integrals to investigate such concepts as areas of surfaces, volumes of geometric solids, and lengths of curves. All the examples we have listed, and many more, will be discussed in detail as we progress through this book. There is literally no end to the applications of calculus. Indeed, in the future perhaps you, the reader, will discover new uses for this important branch of mathematics. The derivative and the definite integral are defined in terms of certain limiting processes. The notion of limit is the initial idea which separates calculus from the more elementary branches of mathematics. Sir Isaac Newton (1642-1727) and Gottfried Wilhelm Leibniz (1646-1716) discovered the connection between derivatives and integrals. Because of this, and their other contributions to the subject, they are credited with the invention of calculus. Many other mathematicians have added a great deal to its develop- ment. The preceding discussion has not answered the question "What is calculus?" Actually, there is no simple answer. Calculus could be called the study of limits, derivatives, and integrals; however, this statement is meaning- less if definitions of the terms are unknown. Although we have given a few examples to illustrate what can be accomplished with derivatives and inte- grals, neither of these concepts has been given any meaning. Defining them will be one of the principal objectives of our early work in this text. 10. Prerequisites for Calculus This chapter contains topics necessary for the study of calculus. After a briefreview ofreal numbers, coordinate systems, and graphs in two dimensions, we turn our J.J Real Numbers attention to one of the most important concepts in mathematics-the notion offunction. Real numbers are used considerably in precalculus mathematics, and we will assume familiarity with the fundamental properties of addition, subtraction, multiplication, division, exponents and radicals. Throughout this chapter, unless otherwise specified, lower-case letters a, b, c, ... denote real numbers. The positive integers 1, 2, 3, 4, ... may be obtained by adding the real number 1successively to itself. The integers consist ofall positive and negative integers together with the real number 0. A rational number is a real number that can be expressed as a quotient ajb, where a and bare integers and b -1= 0. Real numbers that are not rational are called irrational. The ratio of the circumference of a circle to its diameter is irrational. This real number is denoted by n and the notation n ~ 3.1416 is used to indicate that n is ap- proximately equal to 3.1416. Another example ofan irrational number is .)2. Real numbers may be represented by nonterminating decimals. For example, the decimal representation for the rational number 7434/2310 is found by long division to be 3.2181818 ... , where the digits I and 8 repeat indefinitely. Rational numbers may always be represented by repeating decimals. Decimal representations for irrational numbers may also be obtained; however, they are nonterminating and nonrepeating. It is possible to associate real numbers with points on a line lin such a way that to each real number a there corresponds one and only one point, and I 11. 2 1 Prerequisitesfor Calculus (J.J) conversely, to each point Pthere corresponds precisely one real number. Such an association between two sets is referred to as a one-to-one correspondence. We first choose an arbitrary point 0, called the origin, and associate with it the real number 0. Points associated with the integers are then determined by considering successive line segments of equal length on either side of 0 as illustrated in Figure 1.1. The points corresponding to rational numbers such as 253 and -tare obtained by subdividing the equal line segments. Points associated with certain irrational numbers, such as fi, can be found by geometric construction. For other irrational numbers such as rc, no con- struction is possible. However, the point corresponding to n can be approxi- mated to any degree of accuracy by locating successively the points corres- ponding to 3, 3.1, 3.14, 3.141, 3.1415, 3.14159, .... It can be shown that to every irrational number there corresponds a unique point on I and, con- versely, every point that is not associated with a rational number corresponds to an irrational number. 0 B A ~~r 17; ; ; -3 -2 0 4 b a .,fi 1f 1 3 23 -2 2 5 FIGURE 1.1 The number a that is associated with a point A on I is called the coordinate of A. An assignment of coordinates to points on I is called a coordinate system for /, and I is called a coordinate line, or a real line. A direction can be assigned to I by taking the positive direction to the right and the negative direction to the left. The positive direction is noted by placing an arrowhead on I as shown in Figure 1.1. The real numbers which correspond to points to the right of 0 in Figure 1.1 are called positive real numbers, whereas those which correspond to points to the left of 0 are negative real numbers. The real number 0 is neither positive nor negative. The collection of positive real numbers is closed relative to addition and multiplication; that is, if a and b are positive, then so is the sum a + band the product ab. If a and bare real numbers, and a - b is positive, we say that a is greater than h and write a > b. An equivalent statement is h is less than a, written b < a. The symbols > or < are called inequality signs and expressions such as a > b or b < a are called inequalities. From the manner in which we con- structed the coordinate line I in Figure 1.1, we see that if A and B are points with coordinates a and b, respectively, then a > b (orb < a) ifand only ifA lies to the right ofB. Since a - 0 = a, it follows that a > 0 if and only if a is positive. Similarly, a < 0 means that a is negative. The following properties of inequalities can be proved. If a > b and b > c, then a > c. If a > b, then a + c > b + c. If a > b and c > 0, then ac > be. If a > b and c < 0, then ac < be. Analogous properties for" less than" can also be established. 12. 1-41=4 141=4 ,------"-...~ + I I I + I I I + I -5 -4 -3 -2 -1 0 1 2 3 4 5 I FIGURE 1.2 Definition (1.2) (1.3) (1.4) The Triangle Inequality (1.5) Real Numbers 1.1 3 The symbol a ~ b, which is read a is greater than or equal to h, means that either a > bora = b. The symbol a < b < c means that a < band b < c, inwhichcasewesaythathishetweenaandc. Thenotationsa :s; b,a < b :s; c, a :s; b < c, a :s; b :s; c, and so on, have similar meanings. Another property, called completeness, is needed to characterize the real numbers. This property will be discussed in Chapter 11. If a is a real number, then it is the coordinate of some point A on a co- ordinate line /, and the symbol IaIis used to denote the number of units (or distance) between A and the origin, without regard to direction. Referring to Figure 1.2 we see that for the point with coordinate -4 we have I-41 = 4. Similarly, 141 = 4. In general, if a is negative we change its sign to find Ia I, whereas if a is nonnegative then IaI = a. The nonnegative number Ia I is called the absolute value of a. The following definition of absolute value summarizes our remarks. lal = { a -a if a~ 0 if a< 0 Example 1 Find 131, 1-31,101, IJ2- 21, and 12- fll. Solution Since 3, 2 - J2, and 0 are nonnegative, we have 131 = 3, 12 - J21 = 2 - J2, and I0 I = 0. Since -3 and J2 - 2 are negative, we use the formula IaI = -a ofDefinition (1.2) to obtain 1-31 = -( -3) = 3 and IJ2- 21 = - b if and only if a > b or a < - b Ia I = b if and only if a = b or a = -b. It follows from the first and third properties stated in (1.4) that Ia I :s; b if and only if -b :s; a :s; b. Ia + bl ;S; lal + lbl 13. 4 1 Prerequisitesfor Calculus 5= 17-21= 12-71 ~ I I I I + I I I I + I -2 -1 0 1 2 3 4 5 6 7 8 1 FIGURE 1.3 Definition (1.6) A B OC D l l t l l t t l l l l t l . -5 -3 0 1 6 1 FIGURE 1.4 Proof From (1.3), -lal ~a~ lal and -lbl ~ b ~ lbl. Adding corresponding sides we obtain -(lal + lbi) ~a+ b ~ lal + lbl. Using the remark preceding this theorem gives us the desired conclusion. D We shall use the concept of absolute value to define the distance between any two points on a coordinate line. Let us begin by noting that the distance between the points with coordinates 2 and 7 shown in Figure 1.3 equals 5 units on /. This distance is the difference, 7 - 2, obtained by subtracting the smaller coordinate from the larger. If we employ absolute values, then, since 17 - 21 = 12 - 71, it is unnecessary to be concerned about the order of subtraction. We shall use this as our motivation for the next definition. Let a and b be the coordinates of two points A and B, respectively, on a coordinate line 1. The distance between A and B, denoted by d(A,B), is defined by d(A, B) = lb- al. The number d(A, B) is also called the length of the line segment AB. Observe that, since d(B, A)= Ia- bl and lb- al = Ia- bl, we may write d(A, B) = d(B, A). Also note that the distance between the origin 0 and the point A is d(O, A)= Ia- Ol = IaI, which agrees with the geometric interpretation of absolute value illustrated in Figure 1.2. Example 2 IfA, B, C, and D have coordinates -5, -3, 1, and 6, respectively, find d(A, B), d(C, B), d(O, A), and d(C, D). Solution The points are indicated in Figure 1.4. By Definition (1.6), d(A, B) = I- 3- (- 5)I = I- 3+ 51 = 121 = 2. d(C, B)= 1-3- 11 = 1-41 = 4. d(O, A)= l-5- 01 =I-51= 5. d(C, D) = 16 - 11 = 151 = 5. 14. (1.7) ( ) a b ( I ) -1 0 3 ( ) 0 2 4 FIGURE 1.5 Open intervals (a, b), (-I, 3), and (2, 4) (1.8) Real Numbers 1.1 5 The concept ofabsolute value has uses other than that offinding distances between points. Generally, it is employed whenever one is interested in the magnitude or numerical value of a real number without regard to its sign. In order to shorten explanations it is sometimes convenient to use the notation and terminology of sets. A set may be thought of as a collection of objects of some type. The objects are called elements of the set. Throughout our work ~will denote the set of real numbers. If Sis a set, then a E S means that a is an element of S, whereas a S signifies that a is not an element of S. Ifevery element ofa set Sis also an element ofa set T, then Sis called a subset ofT. Two sets Sand Tare said to be equal, written S = T, if Sand Tcontain precisely the same elements. The notation S #- T means that Sand Tare not equal. If Sand Tare sets, their unionS u T consists ofthe elements which are either in S, in T, or in both S and T. The intersection S n T consists of the elements which the sets have in common. Ifthe elements ofa set Shave a certain property, tpen we write S = {x : ...} where the property describing the arbitrary element x is stated in the space after the colon. For example, {x: x > 3} may be used to represent the set of all real numbers greater than 3. Of major importance in calculus are certain subsets of~ called intervals. If a < b, tpe symbol (a, b) is sometimes used for all re!!l numbers between a and b. This set is called an open interval. Thus we have: (a, b) = {x: a < x < b}. The numbers a and b are called the endpoints of the interval. The graphofa set Sofreal numbers is defined as the points on a coordinate line that correspond to the numbers inS. In particular, the graph ofthe open interval (a, b) consists ofall points between the points corresponding to a and b. In Figure 1.5 we have sketched the graphs of a general open interval (a, b) and the special open intervals (- 1, 3) and (2, 4). The parentheses in the figure indicate that the endpoints of the intervals are not to be included. For convenience, we shall use the terms interval and graph of an interval inter- changeably. If we wish to include an endpoint of an interval, a bracket is used instead of a parenthesis. If a < b, then closed intervals, denoted by [a, b], and half- open intervals, denoted by [a, b) or (a, b], are defined as follows. [a, b] = { x: a ~ x ~ b} [a, b) = {x: a ~ x < b} (a, b] = {x: a< x ~ b} Typical graphs are sketched in Figure 1.6, where a bracket indicates that the corresponding endpoint is part of the graph. [ 3 [ ) ( 3a b a b a b FIGURE 1.6 15. 6 1 Prerequisitesfor Calculus (1.9) In future discussions of intervals, whenever the magnitudes of a and b are not stated explicitly it will always be assumed that a < b. Ifan interval is a subset of another interval I it is called a subinterval of/. For example, the closed interval [2, 3] is a subinterval of [0, 5]. We shall sometimes employ the following infinite intervals. (a, oo) = {x: x >a} (-oo,a) = {x:x 2x - 5. Solution The following inequalities are equivalent: 4x + 3 > 2x- 5 4x > 2x- 8 2x > -8 X> -4 Hence the solutions consist of all real numbers greater than -4, that is, the numbers in the infinite interval (- 4, oo). E l 4 I . 1. 4 - 3x xamp e So ve the mequa 1ty -5 < --2- < 1. Solution We may proceed as follows: 4- 3x -5 3 if and only if either 2x- 7 > 3 or 2x- 7 < -3. The first of these two inequalities is equivalent to 2x > 10, or x > 5. The second is equivalent to 2x < 4, or x < 2. Hence the solutions of12x - 71 > 3 are the numbers in the union (- oo, 2) u (5, oo). 1.1 Exercises In Exercises 1 and 2 replace the comma between each pair of real numbers with the appropriate symbol , or =. 1 (a) -2, -5 (c) 6- I, 2 + 3 (e) 2, J4 2 (a) -3,0 (c) 8, -3 (e) Ji, 1.4 (b) -2, 5 (d) !. 0.66 (f) 1!:, (b) -8, -3 (d)i-i,-fs (f) tm. 3.6513 Rewrite the expressions in Exercises 3 and 4 without using symbols for absolute values. 3 (a) 12-51 (b) I-51+ 1-21 (c) 151 + 1-21 (d) 1-51-l-21 .(e) In - 22/11 (f) ( -2)/l-21 (g) It- o.s1 (h) I< -WI (i) 15 - X Iif X > 5 (j) Ia - bIif a < b 4 (a) 14- 81 (b)l3-nl (c) 1-41-1-81 (d) 1-4 + 81 (e) 1-W (f) 12- fil (g) l-0.671 (h) -l-31 (i) lx2 + 11 (j) l-4- x2 1 5 If A, B, and C are points on a coordinate line with co- ordinates -5, -I, and 7, respectively, find the following distances. (a) d(A, B) (c) d(C, B) (b) d(B, C) (d) d(A, C) 6 Rework Exercise 5 if A, B, and C have coordinates 2, -8, and -3, respectively. Solve the inequalities in Exercises 7-34 and express the solutions in terms of intervals. 7 5x- 6 >II 8 3x- 5 < 10 9 2- 7x ~ 16 10 7- 2x ~ -3 11 12x + 11 > 5 12 lx+21 -7 16 5 > 2- 9x > -4 3- 7x 18 0 ~ 4x- I~ 217 -1 0 7- 2x X + 21 lx- 101 < 0.3 22 12x + 31- - , a+ f>), then x is also in (~, ~), and, therefore, .f(x) is in (a2 - , a2 + ). Hence, by Definition (2.4), limx~a x 2 = a2 Although we have considered only a > 0, a similar argument applies if a :s; 0. To shorten explanations, whenever the notation limx~a .f(x) = L is used we shall often assume that all the conditions given in Definition (2.3) are satisfied. Thus, it may not always be pointed out that f is defined on an open interval containing a. Moreover, we shall not always specify L but merely write "limx~a .f(x) exists," or "f(x) has a limit as x approaches a." The phrase "find limx~a f(x)" means "find a number L such that Iimx~a f(x) = L." If no such L exists we write "Iimx~a f(x) does not exist." 67. 58 2 Limits and Continuity ofFunctions y I I I -------- y = L + F -------- y = L - E 1 10 I I I I I I I I f(x) = l X FIGURE 2.9 X It can be proved that if f(x) has a limit as x approaches a, then that limit is unique (see Appendix II). In the following example we return to the function considered at the beginning of this section and prove that the limit exists by means of Definition (2.3). Example 2 Prove that limx~4 !(3x - 1) = 1z'. Solution Let .f(x) = !(3x - 1), a = 4, and L = 'l. According to Defini- tion (2.3) we must show that for every & > 0, there exists a {J > 0 such that if 0 < lx - 41 < b, then 1!(3x - 1) - 'll < &. A clue to the choice of {J can be found by examining the last inequality involving t:. The following is a list of equivalent inequalities. I!(3x - 1) - 1z' 1< & ! I(3x - 1) - 111 < & 13x - 1 - 111 < 2& l3x-121 0 often requires a great deal ofingenuity. In Section 2.3 we shall introduce theorems which can be used to find many limits without resorting to a search for the general number {J that appears in Definition (2.3). The next two examples indicate how the geometric interpretation illustrated in Figure 2.7 may be used to show that certain limits do not exist. Example 3 Show that limx~o 1/x does not exist. Solution true that Let us proceed in an indirect manner. Thus, suppose it were I. 1 Im-= L x~o X for some number L. Let us consider any pair of horizontal lines y = L & as illustrated in Figure 2.9. Since we are assuming that the limit exists, it 68. y I I I I Definition of Limit 2.2 59 should be possible to find an open interval (0 - b, 0 + b) or equivalently, (- b, b), containing 0, such that whenever - b < x < b and x # 0, the point (x, 1/x) on the graph lies between the horizontal lines. However, since 1/x can be made as large as desired by choosing x close to 0, not every point (x, 1/x) with nonzero x-coordinate in (- b, b) has this property. Con- sequently our supposition is false; that is, the limit does not exist. ;:~~:::====lt======}I Example 4 lf.f(x) = lxl/x, show that limx-o .f(x) does not exist. Solution If x > 0, then IxI /x = xjx = 1 and hence, to the right of the y-axis, the graph of .f coincides with the line y = 1. If x < 0, then lxl/x = -x/x = -1, which means that to the left ofthey-axis the graph offcoincides with the line y = -1. If it were true that limx-o f(x) = L for some L, then the preceding remarks imply that -1 ::; L ::; 1. As shown in Figure 2.10, if we consider any pair of horizontal lines y = L e, where 0 < e < 1, then there exist points on the graph which are not between these lines for some nonzero x in every interval ( -b, b) containing 0. It follows that the limit does not exist. -8 I 10 X I I ----..;. I f(x)=~ I X I I FIGURE 2.10 2.2 Exercises Establish the limits in Exercises 1-10 by means of Definition (2.3). lim (5x - 3) = 7 2 lim (2x + 1) = - 5 x~2 x-+-3 3 lim (10 - 9x) = 64 4 lim (8x - 15) = 17 x-+-6 x~4 5 lim - + 2 =-(X ) 13 x~s 4 4 6 lim (9 -::) = 8 x~6 6 7 lim 5 = 5 8 lim 3 = 3 x~3 x~5 9 lim x = n 10 lim 2 = 2 x~a Use the method illustrated in Examples 3 and 4 to show that the limits in Exercises 11-14 do not exist. . lx- 31 12 . X+ 2 11 hm-- hm-- x~3 X- 3 x~-2IX + 21 I I 13 lim-- 14 lim 2 x~ -5 (x + 5) x~l(x-1) 15 Give an example of a functionfwhich is defined at a and for which Iimx~af(x) exists, but Iimx~af(x) =F f(a). 16 Ifjis the greatest integer function and a is any integer, show that Iimx~af(x) does not exist. 17 Letfbe defined by the following conditions:f(x) = 0 if x is rational and f(x) = 1 if x is irrational. Prove that for every real number a, limx~af(x) does not exist. 18 Why is it impossible to investigate limx~oJx by means of Definition (2.3)? Use the graphical technique illustrated in Example I to verify the limits in Exercises 19-22. 20 lim Jx = Ja, a > 0 21 lim~=fa 69. 60 2 Limits and Continuity ofFunctions 2.3 Theorems on Limits Theorem (2.5) Theorem (2.6) It would be an excruciating task to solve each problem on limits by means of Definition (2.3). The purpose of this section is to introduce theorems that may be used to simplify the process. To prove the theorems it is necessary to employ the definition of limit; however, once they are established it will be possible to determine many limits without referring to an a or a b. Several theorems are proved in this section; the remaining proofs may be found in Appendix II. The simplest limit to consider involves the constant function defined by f(x) = c, where cis a real number. If.fis represented geometrically by means of coordinate Jines I and /', then every arrow from I terminates at the same point on /', namely, the point with coordinate c, as indicated in Figure 2.11. ~I [' ) x a z I c - c C + E FIGURE 2.11 It is easy to prove that for every real number a, Jimx-af(x) = c. Thus, if a > 0, consider the open interval (c - a, c + a) on /' as illustrated in the figure. Since.f(x) = cis always in this interval, any number [J will satisfy the conditions of Definition (2.3); that is, for every [J > 0, if 0 < lx- al < b, then l.f(x)- cl 0 there exists a (j > 0 such that if 0 < jx- al < b, then j(mx +b)- (ma + b)i 0, if 0 < jx- al < b, where () = e/lml then the last inequality in the list is true, and hence, so is the first inequality, which is what we wished to prove. D As special cases of Theorem (2.6), we have lim x =a lim (3x - 5) = 3 4 - 5 = 7 lim (l3x + j2) = 13j2 + J2 = 14j2. x---+/2. The next theorem states that if a function f has a positive limit as x approaches a, thenf(x) is positive throughout some open interval containing a, with the possible exception of a. If limx-+a f(x) = L and L > 0, then there exists an open interval (a-(), a+ b)containing a such thatf(x) > Oforallxin(a- b, a+ b), except possibly x = a. Proof lfe = L/2,thentheinterval(L- e, L + e)containsonlypositive numbers. By Definition (2.4) there exists a (j > 0 such that whenever xis in the open interval (a - b, a + {j) and x =1= a, thenf(x) is in (L - e, L + e), and hence f(x) > 0. D In like manner, it can be shown that iffhas a negative limit as x approaches a, then there is an open interval I containing a such that f(x) < 0 for every x in /, with the possible exception of x = a. 71. 62 2 Limits and Continuity ofFunctions Theorem (2.8) Many functions may be expressed as sums, differences, products, and quotients ofother functions. In particular, suppose a functions is a sum oftwo functions f and g, so that s(x) = f(x) + g(x) for every x in the domain of s. If f(x) and g(x) have limits L and M, respectively, as x approaches a, it is natural to conclude that s(x) has the limit L + Mas x approaches a. The fact that this and analogous statements hold for products and quotients are consequences of the next theorem. If lim f(x) = Land lim g(x) = M, then (i) lim [f(x) + g(x)] = L + M. (ii) lim [f(x) g(x)] = L M. (iii) I. [f(x)J L .d1m -) = M' prov1 ed M =1- 0. x~a g(x (iv) lim [cf(x)] = cL. (v) lim [f(x) - g(x)] = L - M. Although the conclusions of Theorem (2.8) appear to be intuitively evident, the proofs are rather technical and require some ingenuity. Proofs for (i)-(iii) may be found in Appendix II. Part (iv) of the theorem follows readily from part (ii) and Theorem (2.5) as follows: lim [cf(x)] = [tim c] [limf(x)] = cL. x-a x-a x-a Finally, to prove (v) we may write f(x)- g(x) = f(x) + (-l)g(x) and then use (i) and (iv) (with c = -I). The conclusions of Theorem (2.8) are often written as follows: (i) lim [f(x) + g(x)] = lim f(x) + lim g(x) (ii) lim [f(x) g(x)] = lim f(x) lim g(x) x~a [ /( )] limf(x) (iii) lim (x) = ;.~a (), if lim g(x) =1- 0 x~a g X 1m g X x~a (iv) ~i~ [cf(x)] = c[~i~f(x)] (v) lim [f(x)- g(x)] = limf(x)- lim g(x). 72. Theorem (2.9) Theorems on Limits 2.3 63 F. d 1. 3x + 4 Example I m 1m - - - . x-+ 2 5x + 7 Solution The numerator and denominator of the indicated quotient define linear functions whose limits exist by Theorem (2.6). Moreover, the limit of the denominator is not 0. Hence by (iii) of Theorem (2.8) and (2.6), 3 4 lim (3x + 4) I. X + x-+2 1m - - - = "-:--'::.....,-::-----,=c- x-+2 5x + 7 lim (5x + 7) x-+2 3(2) + 4 5(2) + 7 10 17. Notice the simple manner in which the limit in Example 1 was found. It would be a lengthy task to verify the limit by means of Definition (2.3). Theorem (2.8) may be extended to limits of sums, differences, products, and quotients that involve any number of functions. An application of (ii) to three (equal) functions is given in the next example. Example 2 Prove that for every real number a, limx....a x3 = a3 Solution Since limx....a x = a we may write lim x3 = lim (x x x) x-+a x-+a = (lim x) (lim x) (lim x) x~a x-+a x-+a The technique used in Example 2 can be extended to xn, where n is any positive integer. We merely write xn as a product x x x of n factors and then take the limit of each factor. This gives us (i) of the next theorem. Part (ii) may be proved in similar fashion by using (ii) of Theorem (2.8). If n is a positive integer, then (i) lim xn = an. x-+a (ii) ~i~ [f(x)]n = [~i~f(x)r provided ~i~ f(x) exists. Example 3 Find limx.... 2 (3x + 4)5 . Solution Applying (ii) of (2.9) and Theorem (2.6), lim (3x + 4)5 = [lim (3x + 4)] 5 x-+2 x-+2 = [3(2) + 4] 5 = 105 = 100,000. 73. 64 2 Limits and Continuity of Functions Theorem (2.10) Corollary (2.11) Example 4 Find limx~ _2 (5x3 + 3x2 - 6). Solution We may proceed as follows (supply reasons): lim (5x3 + 3x2 - 6) = lim (5x3) + lim (3x2) - lim (6) x- -2 x- -2 x--2 x-+-2 x--2 x-+-2 = 5(- 2? + 3(- 2)2 - 6 = 5(-8) + 3(4)- 6 = -34. Note that the limit in Example 4 is the number obtained by substituting -2 for x in 5x3 + 3x2 - 6. The next theorem states that the same is true for the limit of every polynomial. Iff is a polynomial function and a is a real number, then limf(x) = f(a). Proof We may write .f(x) in the form f(x) = bnx" + bn_ 1xn-! + + b0 where the bi are real numbers. As in Example 4, lim f(x) = lim (bnx") + lim (bn_ 1x"- 1) + + lim b0 .---+(1 x -a = bn lim (x") + bn-! lim (x"- 1) ++lim b0 x~a x~a x~a D If q is a rational function and a is in the domain of q, then lim q(x) = q(a). Proof We may write q(x) = f(x)jh(x) where f and h are polynomial functions. If a is in the domain of q, then h(a) =1- 0. Using (iii) of Theorem (2.8) and (2.10), . ~i~ f(x) f(a) ~~~ q(x) = lim h(x) = h(a) = q(a). D 74. Theorem (2.12) Theorems on Limits 2.3 E , 5 F" d I" 5x2 - 2x + 1 xamp.e m xi~ 6x _ 7 Solution Applying Corollary (2.11), I. 5x2 - 2x + 1 1m ---=----=--6x -7x-+3 5(3)2 - 2(3) + 1 6(3)- 7 45- 6 + 1 40 18 - 7 65 The following theorem states that for positive integral roots of x, we may determine a limit by substitution. The proof makes use of the definition of limit and may be found in Appendix II. If a > 0 and n is a positive integer, or if a ::s;; 0 and n is an odd positive integer, then lim ylx = .ja. x-+a If m and n are positive integers and a > 0, then using (ii) of (2.9) and Theorem (2.12), lim (yfxr = (lim v~x)m = (.jar. x-+a x-+a In terms of rational exponents, x-+a This limit formula may be extended to negative exponents by writing x-' = 1/x' and then using (iii) of (2.8). . . x 213 + 3JxExample 6 Fmd !~ 4 _ (16/x) . Solution The reader should supply reasons for each of the following steps. 2/3 3 r: lim (x2i3 + 3jX) I. X + yX x-+8 1m = ~~~---- x-+8 4 - (16/x) lim (4 - (16/x)) x-+8 lim 4 - lim (16/x) x-+8 x-+8 - g2/3 + 3JS - 4- (16/8) = 4 + 6J2 = 2 + 3J24-2 75. 66 y 2 Limits and Continuity ofFunctions Theorem (2.13) The Sandwich Theorem (2.14) Y = f(x) a X FIGURE 2.12 If a function f has a limit as x approaches a, then lim j]Tx) = y/limf(x) x->a provided either n is an odd positive integer or n is an even positive integer and limx_,a f(x) > 0. The preceding theorem will be proved in Section 2.6. In the meantime we shall use it whenever applicable to gain experience in finding limits that involve roots of algebraic expressions. Example 7 Find Iimx~s J3x2 - 4x + 9. Solution Using Theorems (2.13) and (2.10), lim ~hx2 - 4x + 9 = Jlim (3x2 - 4x + 9) . ___,. 5 x-->5 = J75 - 20 + 9 = 164 = 4. The beginning student should not be misled by the preceding examples. It is not always possible to find limits merely by substitution. Sometimes other devices must be employed. The next theorem concerns three functions f, h, and g, where h(x) is always "sandwiched" between f(x) and g(x). If f and g have a common limit L as x approaches a, then as stated below, h must have the same limit. If f(x) ~ h(x) ~ g(x) for all x in an open interval containing a, except possibly at a, and if lim f(x) = L = lim g(x), then lim h(x) = L. x->a A proof of the Sandwich Theorem based on the definition of limit may be found in Appendix II. The result is also clearfrom geometric considerations. Specifically, if f(x) ~ h(x) ~ g(x) for all x in an open interval containing x, then the graph of h lies "between" the graphs off and gin that interval, as illustrated in Figure 2.12. Iff and g have the same limit Las x approaches a, then evidently, h also has the limit L. 76. Theorems on Limits 2.3 67 Example 8 Prove that limx-+o x sin(1/x) = 0. Solution The limit cannot be found by substituting 0 for x, or by using an algebraic manipulation. However, since all values of the sine function are between -1 and 1it follows that if x =f. 0, Isin (1/x)l" :$;; 1and, therefore, Ix sin ~I = Ix II sin~I :$;; Ix 1. Consequently, 0 :$;; Ix sin~ I:$;; lxl. It is not difficult to show that limx-+o IxI = 0. Hence, by the Sandwich Theorem (2.14), with f(x) = 0 and g(x) = IxI, we see that lim Ix sin .!_ I = 0. x-+0 X It now follows from the Definition of Limit (2.3) that 1. . 1 01m xsm- = . x-+0 X 2.3 Exercises In Exercises 1-36 find the limits, if they exist. 1 lim (3x3 - 2x + 7) x-+ -2 3 lim (x2 + 3Xx - 4) x-+,/i 5 lim .jx2 - 5x - 4 x-+4 7 limO 9 lim 15 x-+/2 . 2x2 + 5x- 3 11 hm --::---:2.------::---=- x-+112 6x - 7x + 2 x-2 13 lim-- x-+2 x3 - 8 X- 16 15 lim--- x-+16 Jx- 4 6s- 1 17 lim-- s-+4 2s - 9 19 lim(~- - 1-) x-+1 X- 1 X - 1 2 lim (5x2 - 9x - 8) x-+4 4 lim (3t + 4)(7t - 9) r-+-3 6 lim jx4 - 4x + 1 x-+ -2 4x2 - 6x + 3 8 lim --..,.--.-3 - - - x-112 16x + 8x - 7 10 lim j2 x-+ 15 12 lim x + 3 x-+,- 3 (1/x) + (1/3) x2 - x- 2 14 lim 2 x-+2 (x - 2) x3 + 8 16 lim--- x--2 x4- 16 18 lim (x - 3.1416) 20 lim (Jx + ~)6 x-+1 y X . 2Jx + x312 21 hm _.!...-=-- x-+16 fx + 5 23 r 2 + 5x - 3x3 Im J x2- 1 x-+3 . 4- Ji6+h25 hm h h-+0 v-+3 I. (4t2 + 5t- w31 Im 4 t-+-1 (6t + 5) 33 lim x2 - 81 x-+93- Jx . (2 + h)- 2 - 2- 2 35 hm ---'----- h-o h 16x213 22 lim 413 x-+-8 4- X 26 lim (~)(-1 -1)h-+0 h Ji+h 28 lim (x + 4)3(x - 6)2 x-+6 30 lim j3k2 + 4.j3k + 2 k-+2 32lim~r-+7 (t - 5) 34 I. X- 8 I m - - - x-+8 fx- 2 . (9+h)- 1 -9- 1 36 hm -'----.:.....,.---- h-+0 h 37 If r is any rational number and a > 0, prove that limx-+ax' =a'. Under what conditions will this be true if a < 0? 77. 68 2 Limits and Continuity ofFunctions 38 If limx-af(x) = L =f. 0 and limx-.g(x) = 0, prove that limx-aLf(x)/g(x)] does not exist. (Hint: Assume there is a number M such that limx-aLf(x)/g(x)] = M and consider 41 If c is a nonnegative real number and 0 ::;:; ((x) ::;:; c for every x, prove that limx-o x2 ((x) = 0. 42 Provethatlimx_0 x4 sin(l/fx) = 0. (Hint: Sec Example 8.) 43 Prove that iff has a negative limit as x approaches a, then there is some open interval I containing a such thatf(x) is negative for every x in I except possibly x = a. lim f(x) = lim [g(x) f(x)]. x-a x-a g(x) 39 Use the Sandwich Theorem and the fact that limx-o(lxl + I)= I to prove that limx-o (x2 + I)= I. 40 Use the Sandwich Theorem withf(x) = 0 and g(x) = Ixi to prove that lim lxl = 0. x-oJx4 + 4x2 + 7 2.4 One-Sided Limits Definition (2.15) If f(x) = ~and a > 2, then .f is defined throughout an open interval containing a and, by Theorem (2.13), lim~ = Jlim (x - 2) = J a - 2. The case a = 2is not covered by Definition (2.3) since there is no open interval containing 2 throughout which/ is defined (note that v x - 2 is not real if x < 2). A natural way to extend the definition of limit to include this excep- tional case is to restrict x to values greater than 2. Thus, we replace the con- dition 2 - b < x < 2 + b, which arises from Definition (2.3), by the condition 2 < x < 2 + b. The corresponding limit is called the limit off(x) as x approaches 2 from the right, or the right-hand limit of .j"X=2 as x approaches 2. Let .f be a function that is defined on an open interval (a, c), and let L be a real number. The statement lim f(x) = L x-a+ means that for every e > 0, there exists o> 0, such that if a 0, there exists a b > 0, such that if a - b < x < a, then l.f(x) - L I < B. Iflimx~u-f'(x) = L, we say that the limit ofj(x) as x approaches a from the left, is L; or that Lis the left-hand limit off(x) as x approaches a. The symbol x -+ a- is used to indicate that xis restricted to values less than a. A geometric illustration of Definition (2.16) is given in (ii) of Figure 2.13. Note that for the l~(t-hand limit, xis in the left half (a - b, a) of the interval (a - b, a + b). Sometimes Definitions (2.15) and (2.16) are referred to as one-sided limits of.f(x) as x approaches a. The relation between one-sided limits and limits is stated in the next theorem. The proof is left as an exercise. If.fis defined throughout an open interval containing a, except possibly at a itself, then limx~a.f(x) = L if and only if both limx--+a- .f(x) = L and limx~a+ .f(x) = L. 79. 70 y 2 Limits and Continuity of Functions X FIGURE 2.14 y X FIGURE 2.15 X FIGURE 2.16 The preceding theorem tells us that the limit of f(x) as x approaches a exists if and only if both the right- and left-hand limits exist and are equal. Theorems similar to the limit theorems of the previous section can be proved for one-sided limits. For example, lim [f(x) + g(x)] = lim f(x) + lim g(x) and lim .fJW = .::/lim f(x) x-a+ x-a+ with the usual restrictions on the existence of limits and nth roots. Analogous results are true for left-hand limits. Example 1 Fine! limx~z +(1 + ~). Solution Using (one-sided) limit theorems, lim (1 + ~) = lim 1 + lim ~ x- 2 + x- 2 + x- 2 + = 1 + J lim (x - 2) x-z+ =1+0=1. The graph of f(x) = 1 + ~is sketched in Figure 2.14. Note that there is no left-hand limit, since~ is not a real number if x < 2. Example 2 Suppose f(x) = IxI /x if x =f. 0 and f(O) = l. Find limx~o+ f(x) and limx~o- f(x). What is limx~of(x)? Solution If x > 0, then IxI = x and f(x) = xjx = l. Consequently, lim f(x) = lim l = l. x-o+ If x < 0, then lxl = -x and f(x) = -xjx = -l. Therefore, lim f(x) = lim (-1) = -l. x-o- x-o- Since these right- and left-hand limits are unequal, it follows from Theorem (2.17) that limx~o f(x) does not exist. The graph off is sketched in Figure 2.15. Example3 Supposef(x) = x + 3ifx =f. landf(1) = 2.Findlimx~ 1 -/(x), limx~ 1 + f(x), and limx~ 1 f(x). Solution The graph off consists of the point P(l, 2) and all points on the line y = x + 3 except the point with coordinates (1, 4) as shown in Figure 2.16. Evidently, limx~ 1 + f(x) = 4 = limx~ 1 - f(x). Hence by Theorem (2.17), limx~ 1 f(x). = 4. Note that limx~ 1 f(x) =f. f(l ). 80. 2.4 Exercises In Exercises 1-22, find the limits, if they exist. (a) lim~ x-+5- (b) lim~ x-+5 + (c) lim~ 3 (a) lim Jx3 - I x-+ 1- (b) lim p=i" x-+ 1 + (c) lim Jx3 - 1 x-1 . lx- 41 5 (a) hm - - x-4- X- 4 (c) lim lx- 41 x-4 X- 4 7 lim (4 + yfx) x-o+ 9 lim cfi+6 + x) x-+-6+ 11 lim (jx2 - 25 + 3) 13 15 17 x-+ 5 + 4 - x2 lim-- x-2 2-x 1 + Jh-=-10lim x-s X+ 3 19 lim x + 7 x--7+ lx + 71 21 . 1 hm - x-+0 + X 2 (a) lim )8=7 x-+2- (b) lim )8=7 x-+2 + (c) limj8=7 4 (a) lim x213 x-+- 8 + (b) lim x213 x- -8- (c) lim x213 x-+-8 x+5 6 (a) lim - - x--s lx +51 , X+ 5 (b) hm - - x-- s- lx + 51 x+5 (c) lim-- x--slx +51 8 lim (4x312 - Jx + 3) x-o+ x-+ 5/2- 12 limx~ x-+3- 14 lim 2x2 + 5x- 12 x--4 x2 +3x-4 . X+ 10 16 hm x--10- j(x + 10)2 ~18 lim x-4 X+ 4 20 22 . In- xl hm --- x-+1[- X- n . 1 hm-- x-s- X - 8 One-Sided Limits 2.4 71 For each f defined in Exercises 23 and 24, find limx-2- f(x), limx-2 f(x), and sketch the graph off. 23 _ {3x if x :::;; 2 f(x)- 2 'f 2X I X> { x3 if x :::;; 2 24 f(x) = 4 - 2x if x > 2 In Exercises 25 and 26 find limx__ 3 f(x), limx-- 3- f(x), and limx__ 3f(x), if they exist. x _ {1/(2- 3x) ifx < -3 25 j( ) - Jx+2 if X;::: -3 { 9/x2 if x < -3 26 j(x) = 4 + X if X: -3 In Exercises 27-29, n denotes an arbitrary integer. For each function f, sketch the graph off and find limx_n_ f(x) and limx-n f(x). 27 f(x) = (-1)" ifn:::;; x < n + 1 f {0 if X = n {X if X = n 28 (x) = 29 f(x) = 01 if x # n if x # n In Exercises 30 and 31, [ ] denotes the greatest integer function and n is an arbitrary integer. 30 Find limx_n_ [x] and limx-n [x]. 31 lfj(x) = X - [x], find limx_n_ f(x) and limx-n f(x). 32 If f(x) = (x2 - 9)/(x - 3) for x # 3 and f(3) = 5, find limx-3 f(x), limx-3- f(x), and limx-3 f(x). 33 Iff is a polynomial function, prove that limx-a f(x) = f(a) and limx_a_ f(x) = f(a) for every real number a. 34 Prove Theorem (2.17). 35 Sketch geometric interpretations for right-hand limits which are analogous to those in Figures 2.5 and 2.6. Do the same for left-hand limits. 81. 72 2 Limits and Continuity ofFunctions 2.5 Limits of Trigonometric Functions Theorem (2.18) y X FIGURE 2.17 Corollary (2.19) Whenever we discuss limits of trigonometric expressions involving sin t, cos x, tan 0, etc., we shall assume that each variable represents a real number or the radian measure of an angle. The following result is important for future developments. lim sin t = 0 Proof Let us first prove that lim1-+o+ sin t = 0. Since we are only interested in positive values of t near zero, there is no loss of generality in assuming that 0 < t < n/2. Let U be the circle of radius 1 with center at the origin of a rectangular coordinate system, and let A be the point (1, 0). If, as illustrated in Figure 2.17, P(x, y) is the point on U that corresponds to t, then by (1.26) the radian measure ofangle AOP is t. Referring to the figure we see that O--> cos to t If - n/2 < t < 0, then 0 < - t < n/2, and from the result just established, sin(-t) l > > cos (- t)o -t Since sin (- t) = -sin t and cos (- t) = cos t, this inequality reduces to (*)o This shows that (*) is also true if - n/2 < t < 0, and hence is true for every tin the open interval (- n/2, n/2) except t = 00 Since lim,~o cost = l, and (sin t)/t is always between cost and l, it follows from the Sandwich Theorem that 0 Roughly speaking, Theorem (2021) implies that if t is close to 0, then (sin t)/t is close to 10 Another way of stating this is to write sin t :::::: t for small values of to It is important to remember that if t denotes an angle, then radian measure must be used in Theorem (2021) and in the approximation formula sin t :::::: to To illustrate, trigonometric tables or a calculator show that to five decimal places, Proof sin (0006) :::::: 0005996 sin (0005) :::::: 0004998 sin (0004) :::::: 0003999 sin (0003) :::::: 00030000 10 l - cost 0 Ill = t We may change the form of (l - cos t)jt as follows: l - cos t l - cos t l + cos t l +cost l - cos2 t t(l + cost) sin2 t t(l + cost) sin t sin t l + cost 0 84. Limits of Trigonometric Functions 2.5 75 Consequently, Im = Im ------I. 1 - cos t 1. (sin t sin t ) r-0 t r-o t 1 + cos t (1. sin t)(I sin t )= 1m-- Im---- r-o t r-o 1 + cost = 1 c~ 1) = 10 = 0. 0 Example I Find limx-o (sin 5x)j2x. Solution We cannot apply Theorem (2.21) directly, since the given expression is not in the form (sin t)jt. However, we may introduce this form (with t = 5x) by using the following algebraic manipulation: I. sin 5x . 1 sin 5x Im--=hm--- x-o 2x x-0 2 X = lim~ sin 5x x-o 2 5x = ~ lim sin 5x . 2 x-o 5x It follows from the definition of limit that x --+ 0 may be replaced by 5x --+ 0. Hence, by Theorem (2.21), with t = 5x, we see that lim sin 5x = ~ (1) = ~ . x-o 2x 2 2 Warning: When working problems of this type, remember that sin 5x =I= 5 sin x. Example 2 Find lim1_ 0 (tan t)/2t. Solution Using the fact that tan t = sin tjcos t, r tan t r (1 sin t 1)~~~2t= ~~~ 2-t-cost = t. 1 . 1 = t. Example 3 Find limx-o (2x + 1- cos x)j3x. Solution We plan to use Theorem (2.22). With this in mind we begin by isolating the part ofthe quotient that involves (1 - cos x)jx and then proceed as follows. I. 2x + 1- cos x 1. (2x 1- cos x)1m = 1m - + ---- x-o 3x x-o 3x 3x = lim (2x) + lim ! (1 - cos x) x-o 3x x-0 3 X 10 2 1 I' 1 - cos X = Im- +- Im---- x-0 3 3 x-0 X =i+tO=i 85. 76 2 Limits and Continuity of Functions 2.5 Exercises Find the limits in Exercises 1-26. 17 lim I -cost . sin (x/2) 18 hm--- 3 5 7 9 ll l3 15 . X sin x hm-.- 2 lim-- x-o Sin X x-o .y;. sin3 t lim 30 +sin 0 lim-- 4 r-o (21)3 o-o o lim 2 +sin x 6 lim 1 - cos 3t x-o 3+x r-o lim 2 cos 0- 2 lim x2 +I 8 o-o 30 x-o X+ COS X sin (-3x) x sin x lim lO lim-- x-o 4x x-o x2 + I 1-cosx 12 lim I - 2x2 - 2 cos x + cos2 x lim x213 x2x-o x-o 4t2 + 3t sin t X COS X- X2 lim 14 lim r-o 12 x-o 2x lim cost lim sin t 16 ,_ 0 I - sin t r-o I +COSt r-o sin t x-0 X lim x +tan x sin2 2t 19 20 lim--2- x-o sin x r-o t 21 lim x cot x 22 esc 2x lim-- x-o x-o COt X ~ lim IX2 csc2 IX 24 I' sin 3x 1m-- -0 x-o sin Sx 25 lim cos (v + n/2) 26 lim sin2 (x/2) v-o v x-o Sin X Establish the limits in Exercises 27-30, where a and bare any nonzero real numbers. 27 . sin ax a 28 I -cos ax hm--=- lim =0 x-o bx b x-o bx 29 . sin ax a hm--=- x-o sin bx b 30 . cos ax hm--= I x-o COS bx 2.6 Continuous Functions Definition (2.23) In arriving at the definition of limx-a f(x) we emphasized the restriction x =1 a. Several examples in preceding sections have brought out the fact that limx-a f(x) may exist even though f is undefined at a. Let us now turn our attention to the case in which a is in the domain off Iff is defined at a and limx-a f(x) exists, then this limit may, or may not, equalf(a). Iflimx_af(x) = f(a) then .fis said to be continuous at a according to the next definition. A function( is continuous at a number a if the following three conditions are satisfied. (i) f is defined on an open interval containing a. (ii) lim f(x) exists. (iii) lim f(x) = f(a). Iff is not continuous at a, then we say it is discontinuous at a, or has a discontinuity at a. 86. y y=f(a)+E f(a) a-6 FIGURE 2.20 Continuous Functions 2.6 77 If .f is continuous at a, then by (i) of Definition (2.23) there is a point (a, .f(a)) on the graph of .f. Moreover, since limx-a.f(x) = .f(a), the closer x is to a, the closer .f(x) is to .f(a) or, in geometric terms, the closer the point (x, .f(x)) on the graph of.fis to the point (a, .f(a)). More precisely, as illustrated in Figure 2.20, for every pair of horizontal lines y = .f(a) s, there exist vertical lines x = a c:5, such that if a - c:5 < x < a + c:5, then the point (x, .f(x)) on the graph of.flies within the shaded rectangular region. Functions that are continuous at every number in a given interval are sometimes thought of as functions whose graphs can be sketched without x lifting the pencil from the paper; that is, there are no breaks in the graph. Another interpretation of a continuous function .f is that a small change in x produces only a small change in the functional value .f(x). These are not accurate descriptions, but rather devices to help develop an intuitive feeling for continuous functions. Example I (a) Prove that the sine function is continuous at 0. (b) Prove that a polynomial function is continuous at every real number a. (c) Prove that a rational function is continuous at every real number in its domain. Solution (a) We must verify the three conditions of Definition (2.23). Since sin x is defined for every x, it is defined in an open interval containing 0, and consequently (i) of(2.23) is fulfilled. Moreover, by Theorem (2.18), lim sin x = 0 x-o and hence condition (ii) of (2.23) holds. Finally, since sin 0 = 0 we have lim sin x = sin 0 x-o and therefore condition (iii) is true. This completes the proof. (b) Apolynomial function .fis defined throughout IR. Moreover, by Theorem (2.1 0), Iimx-a .f(x) = .f(a) for every real number a. Thus .fsatisfies conditions (i)-(iii) of Definition (2.23) and hence is continuous at a. (c) If q is a rational function, then q = .fjh, where .f and h are polynomial functions. Consequently q is defined for all real numbers except the zeros of h. It follows that if h(a) i= 0, then q is defined throughout an open interval containing a. Moreover, by (2.11), limx-a q(x) = q(a). Applying Definition (2.23), q is continuous at a. Since the notion of continuity involves the fact that limx-a .f(x) = .f(a), the following result may be obtained by replacing Lin Definition (2.3) by.f(a). 87. 78 2 Limits and Continuity ofFunctions Theorem (2.24) If a function .f is defined on an open interval containing a, then f is continuous at a if for every e > 0, there exists a o> 0, such that if lx- al < o, then l.f(x)- .f(a)l 1 88. Definition (2.25) y -3 3 X FIGURE 2.22 Continuous Functions 2.6 79 condition (iii) of Definition (2.23) is not satisfied, and hence f has a dis- continuity at x = 1. If[(x) = 1/x, then f has a discontinuity at x = 0. In this case none of the conditions of Definition (2.23) is satisfied (see Figure 2.9). The functions whose graphs are sketched in Figure 2.21 appear to be continuous at numbers other than a. Most functions considered in calculus are of this type; that is, they may be discontinuous at certain numbers of their domains and continuous elsewhere. Ifa functionjis continuous at every number in an open interval (a, b), we say thatfis continuous on the interval (a, b). Similarly, a function is said to be continuous on an infinite interval of the form (a, w) or (- w, b) if it is continuous at every number in the interval. The next definition covers the case of a closed interval. Let a functionfbe defined on a closed interval [a, b]. The function/is continuous on [a, b] if it is continuous on (a, b) and if, in addition, lim f(x) = f(a) and lim f(x) = f(b). If a function f has either a right-hand or left-hand limit of the type indicated in Definition (2.25) we say thatfis continuous from the right at a or that/is continuous from the left at b, respectively. Example 2 If f(x) =~'sketch the graph off and prove that f is continuous on the closed interval [- 3, 3]. Solution By (1.13), the graph ofx2 + y2 = 9 or equivalently y2 = 9 - x2 is a circle with center at the origin and radius 3. It follows that the graph of y = ~ and, therefore, the graph of J, is the upper half of that circle (see Figure 2.22). If -3 < c < 3 then, using Theorem (2.13), limf(x) =lim~=~= f(c). Hence, by Definition (2.23), f is continuous at c. According to Definition (2.25), all that remains is to check the endpoints of the interval using one-sided limits. Since lim f(x) = lim ~ = J9=9 = 0 = f( -3), x--3+ x-+ -J+ f is continuous from the right at -3. We also have lim f(x) = lim J9 - x2 = J9=9 = 0 = f(3) x-+3- x-3- and hence f is continuous from the left at 3. This completes the proof that f is continuous on [- 3, 3]. 89. 80 2 Limits and Continuity ofFunctions Theorem (2.26) It should be evident how to define continuity on other types of intervals. For example, a function .f is said to be continuous on [a, b) or [a, x:;) if it is continuous at every number greater than a in the interval and if, in addition, .fis continuous from the right at a. For intervals of the form (a, b] or (- x:;, b] we require continuity at every number less than b in the interval and also continuity from the left at b. As illustrated in the next example, when asked to discuss the continuity of a function .f we shall list the largest intervals on which f is continuous. Ofcourse .fwill also be continuous on any subinterval of those intervals. Example 3 Discuss the continuity off if f(x) = ~/(x - 4). Solution The function is undefined if the denominator x - 4 is zero (that is, ifx = 4) or if the radicand x 2 - 9 is negative (that is, if- 3 < x < 3). Any other real number is in one of the intervals (- x, - 3], [3, 4), or (4, XJ ). The proof that .f is continuous on each of these intervals is similar to that given in the solution of Example 2. Thus to prove continuity on [3, 4) it is necessary to show that limf(x) = .f(c) if3 < c < 4 and also that lim .f(x) = .f(3). x-3+ We shall leave the details of the proof for this, and the other intervals, to the reader. Limit theorems discussed in Section 2.3 may be used to establish the following important theorem. If the functions .f and g are continuous at a, then so are the sum .f + g, the difference .f - g, the product .fg, and, if g(a) "I= 0, the quotient.f/g. Proof If .f and g are continuous at a, then limx_a.f(x) = .f(a) and limx-a g(x) = g(a). By the definition of sum (.f + g)(x) = .f(x) + g(x). Consequently, lim (.f + g)(x) = lim [.f(x) + g(x)] = lim .f(x) + lim g(x) = .f(a) + g(a) = (.f + g)(a). This proves that .f + g is continuous at a. The remainder of the theorem i~ proved in similar fashion. c 90. Theorem (2.27) (i) (ii) (iii) Continuous Functions 2.6 8/ Iff and g are continuous on an interval I it follows that f + g, f - g, and fg are continuous on I. If, in addition, g(a) i= 0 throughout I, then fjg is continuous on I. These results may be extended to more than two functions; that is, sums, differences, products, or quotients involving any number of continuous functions are continuous (provided zero denominators do not occur). The next result on limits of composite functions has many applications. If .f and g are functions such that limx-a g(x) = b, and if .f is con- tinuous at b, then Proof As was pointed out in Chapter 1(see Figure 1.34), the composite function .f(g(x)) may be represented geometrically by means of three real lines /, 1', and I" as shown in Figure 2.23, where to each coordinate x on I there corresponds the coordinate g(x) on I' and then, in turn, .f(g(x)) on /". We wish to prove that f(g(x)) has the limit .f(b) as x approaches a. In terms of Definition (2.3) we must show that for every e > 0 there exists a lJ > 0 such that if 0 < lx- al < lJ, then lf(g(x))- .f(b)l 0 such that if lz- bl < b1, then l.f(z)- .f(b)l 0 such that if 0 < lx- al < 15, then ig(x)- bl < 151 ( a-o x a a+o ~Ib- o1 g(x) b FIGURE 2.25 Finally, combining (iv) and (iii) we see that ) b + o1 if 0 < lx- al < 15 then I f(g(x))- f(b)l < e which is the desired conclusion (i). [' 0 The principal use of Theorem (2.27) is to establish other theorems. To illustrate, if n is a positive integer and f(x) = ..::(X, then and f(g(x)) = yfgW 1(~i~ g(x)) = ~~i~ g(x). If we now use the fact that the result stated in Theorem (2.13) is obtained, that is, lim yfgW = ~lim g(x), x-+a x-+a where it is assumed that the indicated nth roots exist. The next result follows directly from Theorem 2.27. Ifg is continuous at a and f is continuous at b = g(a), then ~i~ f(g(x)) = 1(~i~ g(x)) = f(g(a)). The preceding theorem states that the composite function off by g is continuous at a. This result may be extended to functions that are continuous on intervals. Sometimes this is expressed by the statement "the composite function of a continuous function by a continuous function is continuous." 92. f(b) w f(a) The Intermediate Value Theorem (2.29) y a c b X FIGURE 2.26 Continuous Functions 2.6 83 Example 4 Iff(x) = IxI, prove that f is continuous at every real number a. Solution Since IxI = JXi we have, by (2.13) a"ud (2.9), lim f(x) = lim lxl = lim JXix-+a x-+a x-+a = Jlim x2 = P = lal = f(a). x-+a Hence, from Definition (2.23), f is continuous at a. A proof of the following important property ofcontinuous functions may be found in more advanced texts on calculus. lfafunctionfis continuous on a closed interval [a, b] and ifj(a) f(b), then f takes on every value between f(a) and f(b) in the interval [a, b]. Theorem (2.29) states that ifw is any number between f(a) and f(b), then there is a number c between a and b such that J(c) = w. If the graph of the continuous function f is regarded as extending in an unbroken manner from the point (a, f(a)) to the point (b, f(b)), as illustrated in Figure 2.26, then for any number w between f(a) and f(b) it appears that a horizontal line with y-intercept w should intersect the graph in at least one point P. The x-co- ordinate c ofPis a number such that f(c) = w. Example 5 Verify the Intermediate Value Theorem (2.29) ifj(x) = Jx + 1 and the interval is [3, 24]. Solution Thefunctionfis continuous on [3, 24] andf(3) = 2,J(24) = 5. Ifw is any real number between 2 and 5we must find a number cin the interval [3, 24] such that f(c) = w, that is, Jc+l = w. Squaring both sides of the last equation and solving for c we obtain c = w2 - 1. This number cis in the interval [3, 24], since if 2 < w < 5, then 4 < w2 < 25, or 3 < w2 - 1 < 24. To check our work we see that f(c) = f(w2 - 1) = j(w2 - 1) + 1 = w. A corollary of Theorem (2.29) is that iff(a) and f(b) have opposite signs, then there is a number c between a and b such that f(c) = 0; that is, f has a zero at c. Geometrically, this implies that ifthe point (a, f(a))on the graph ofa continuous function lies below the x-axis, and the point (b, f(b)) lies above the x-axis, or vice versa, then the graph crosses the x-axis at some point (c, 0), where a < c < b. 93. 84 2 Limits and Continuity of Functions Theorem (2.30) 2.6 Exercises A useful consequence of the Intermediate Value Theorem is the following, where the interval referred to may be either closed, open, half-open, or infinite. Ifa function f is continuous and has no zeros on an interval, then either f(x) > 0 or f(x) < 0 for all x in the interval. Proof The conclusion of the theorem states that under the given hypothesis,.f(x) has the same sign throughout the interval. If this conclusion were false, then there would exist numbers x 1 and x 2 in the interval such that f(x 1) > 0 and f(x 2 ) < 0. By our preceding remarks this, in turn, would imply that f(c) = 0 for some number c between x 1 and x 2 , contrary to hypothesis. Thus, the conclusion must be true. D In Exercises 1-12 show that the function/ is continuous at the given number a. I 15 f(x) = - 2; (0, x) X f(x) = fo=-5 + 3x, a = 4 2 f(x) = 3x2 + 7- 1/Fx.a = -2 3 f(x) = xj(x2 - 4), a = 3 4 f(x) = ljx, a = 10- 6 5 f(x) = Jx 2 + 2, a= -5 6 f(x) = Jx/(2x + 1), a= 8 7 f(x) =cos x, a= 0 (Hint: See Corollary (2.19).) 8 f(x) = sin x +cos x 2 , a= 0 (Hint: See Example x + 3x +I . Exercise 7.) 9 f(x) = sec x, a = 0 10 f(x) = tan x, a = 0 11 f(x) = (cos x)/(1 + sin x), a = 0 12 f(x) = I -sin x - - - a=O I +cos x' I and In Exercises 13-16, show that f is continuous on the indicated interval. 13 f(x) = ~; [4, 8] 14 f(x) = ~; ( -oo, 16] I 16 f(x) = - - ; (1, 3) x-1 In Exercises 17-28 find all numbers for which the function{' is continuous. 3x - 5 17 f(x)=2x2-x-3 x2 - 9 18 f(x) = - - - x-3 19 f(x) = flx-=3 + x 2 X 20 f(x) = Jx-=---4x-4 x-1 21 f(x) = Jx2=1X -I X 22 f(x) = jl-=7 1-x 23 lx + 91 f(x)=~ X 24 f(x) = x2 + 1 94. 5 25 f(x) = - 3- -2 X -X 4x- 7 26 f(x) = (x + 3)(x2 + 2x - 8) J>?=9)25 - x2 27 f(x) = x _ 4 ~28 f(x) = r::-L ..,;x-6 29-34 Discuss the discontinuities of the functions defined in Exercises 27-32 of Section 2.4. { cx2 - 3 if x < 2 35 Suppose f(x) = 2 .f - 2. ex+ 1 x > Find a value of c such thatfis continuous on 1R. { c2x ifx 0, prove that.f(x) is positive throughout an interval containing c. (Hint: See Theorem (2.17).) In Exercises 49-52, verify the Intermediate Value Theorem (2.29) for .f on the stated interval [a, b] by showing that if w is any number between.f(a) andf(b), then there is a number c in [a, b] such that.f(c) = w. 49 f(x) = x3 + I; [-I, 2] 50 f(x)= -x3 ;[0,2] 51 .f(x) = x2 + 4x + 4; [0, I] 52 .f(x) = x2 - x; [-I, 3] 53 If.f(x) = x3 - 5x2 + 7x - 9, use the Intermediate Value Theorem (2.29) to prove that there is a real number a such thatf(a) = 100. 54 Prove that the equation x5 - 3x4 - 2x3 - x + I = 0 has a solution between 0 and I. 95. 86 2 Limits and Continuity ofFunctions 2. 7 Review: Concepts Define or discuss each of the following. 1 The limit of a function as x approaches a 2 The geometric interpretations of limx-a f(x) = L 3 Theorems on limits 4 Limits of polynomial and rational functions 5 Right- and left-hand limits of functions Review: Exercises In Exercises 1-26 find the limit, if it exists. 5x +II 6- 7x lim--- 2 lim 4 x-3Fx+J x--2 (3 + 2x) 3 lim (2x- }4x2 + x) 4 lim(x-~) x---+ -2 x-4- 2x2 + x- 6 lim 3x2 - x- 10 5 lim 2 6 x 2 - x- 2x-3!2 4x - 4x - 3 x-2 x4 - 16 lim7 lim .2 8 x~ 2 X - X - 2 x-3X-3 I (1/x)- (1/5) 9 lim- 10 lim x-o Jx x-5 x-5 8x3 - I 11 lim--- 12 lim 5 x-I/2 2x- I x-2 3-x 0 Jx-fi13 lim--- 14 hm x-3 13- xi x-2 x-2 15 lim (a+ h)4 - a4 16 lim x+3 3--- h-0 h x-+- 3 x3 + 27 17 lim (2 + h)- 3 - 2- 3 18 lim (x2 + 3)0 h-0 h x-5 x2 lim x 2 + sin2 x 19 lim-.- 20 4x2x-0 Sin X x-o sin2 x + sin 2x 2- COS X 21 lim 22 lim x-0 3x x-0 I + sin X 23 lim 2 cos x + 3x- 2 24 lim 3x + I - cos2 x x-0 5x x-o sin x x sin x cos X- I 25 lim 26 lim x-o I -COS X x-o 2x 6 Limits of trigonometric functions 7 Continuous function 8 Discontinuities of a function 9 Continuity on an interval 10 The Intermediate Value Theorem Find the limits in Exercises 27 and 28, where [ ] denotes the greatest integer function. 27 lim ([x] - x2) 28 lim ([x] - x2) x-+ 3 + x-+3- 29 Prove, directly from the definition of limit, that limx_6 (5x- 21) = 9. 30 Suppose .f(x) = I if x is rational and .f(x) = -I if x is irrational. Prove that limx-a.f(x) does not exist for any real number a. In Exercises 31-34, find all numbers for which .fis continuous. 31 f(x) = 2x4 - .y;. + I 32 f(x) = j(2 + x)(3- x) 33 .f(x) = ~ 34 Jx x4 - 16 f(x) = -z-1X - In Exercises 35-38 find the discontinuities off. lx2 - 161 I 35 .f(x) = x 2 - 16 36 .f(x) = x 2 - 16 x 2 - x- 2 x+2 37 .f(x)=--- 38 f(x) = -3-8 x 2 - 2x X - In Exercises 39 and 40, prove that the function.fis continuous at 0. 4 +sin x 39 .f(x) = xz + 2x + 5 40 .f(x) = (2x2 + 3) cos x 41 If .f(x) = ljx 2 , verify the Intermediate Value Theorem (2.29) for.fon the interval [2, 3]. 42 Prove that the equation x 5 + 7x2 - 3x - 5 = 0 has a root between - 2 and - I. 96. The Derivative The derivative ofa function is one of the most powerful tools in mathematics. Indeed, it is indispensable for nonelementary investigations in both the natural and human sciences. We shall begin our work by reformu- lating the notion of tangent line introduced in the pre- ceding chapter and then discussing the velocity of a moving object. These two concepts serve to motivate the definition of derivative given in Section 3.2. The remainder of the chapter is concerned primarily with properties of the derivative. J.J Introduction Definition (3 .1) Let P(a, f(a)) be any point on the graph of a function f Another point on the graph may be denoted by Q(a + h, f(a + h)), where h is the difference between the x-coordinates of Q and P (see (i) of Figure 3.1). By Definition (1.14), the slope mPQ of the secant line through P and Qis f(a + h) - f(a) mPQ = h . In Chapter 2 the slope m of the tangent line l at P was introduced as the limiting value of mPQ as Q approaches P (see (2.2) and (ii) of Figure 3.1). If f is continuous, then we can make Q approach P by letting h approach 0. It is natural, therefore, to define m as follows. Let f be a function that is defined on an open interval containing a. The slope m of the tangent line to the graph offat the point P(a,f(a)) is given by m = limf(a +h) - f(a) h.... O h provided the limit exists. 87 97. 88 3 The Derivative (3.2) y a X a X (i) Slope of secant line: (ii) Slope of tangent line /: I. f(a + h) - f(a) m= I I D - - - - - h f(a + h) - f(a) mPQ = h FIGURE 3.1 If a tangent line is vertical, its slope is undefined and the limit in Defini- tion (3.1) does not exist. Vertical tangent lines will be studied in the next chapter. Example 1 If f(x) = x 2, find the slope of the tangent line to the graph off at the point P(a, a2 ). Solution This problem is the same as that stated for equations in Example 1ofSection 2.1 (see Figure 2.3). Using the quotient in Definition (3.1), f(a + h) - f(a) h (a+ h?- a2 h a2 + 2ah + h2 - a2 h 2ah + h2 h = 2a +h. The slope m of the tangent line is, therefore, m = lim (2a + h) = 2a. h~o One of the main reasons for the invention of calculus was the need for a way to study objects in motion. Let us consider the problem of arriving at a satisfactory definition for the velocity, or speed, of an object at a given instant. We shall assume, for simplicity, that the object is moving on a line. Motion on a line is called rectilinear motion. It is easy to define the average velocity r during an interval of time. We merely use the following formula: Average velocity = ,. = ~ t 98. A B 80 mi 2: 30 P.M. A B 84 mi 2:35 P.M. FIGURE 3.2 Introduction 3.1 89 where t denotes the length of the time interval and dis the distance between the initial position of the object and its position after t units of time. As an elementary illustration, suppose an automobile leaves city A at 1:00 P.M. and travels along a straight highway, arriving at city B, 150 miles from A, at 4:00P.M. Employing (3.2) we see that its average velocity r during the indicated time interval is 150/3, or 50 miles per hour. This is the velocity that, if maintained for three hours, would enable the automobile to travel the distance from A to B. The average velocity gives no information whatsoever about the velocity at any instant. For example, at 2:30 P.M. the automobile's speedometer may have registered 40, or 30, or the automobile may not even have been moving. If we wish to determine the rate at which the automobile is traveling at 2:30P.M., information is needed about its motion or position near this time. For example, suppose at 2:30P.M. the automobile is 80 miles from A and at 2:35 P.M. it is 84 miles from A, as illustrated in Figure 3.2. For the interval from 2: 30 P.M. to 2: 35 P.M. the elapsed time t is 5 minutes, or 1/12 hour, and the distance d is 4 miles. Substituting in (3.2), the average velocity r during this time interval is 4 . r = ( 1112) = 48 miles per hour. However, this is still not an accurate indication of the velocity at 2:30P.M. since, for example, the automobile may have been traveling very slowly at 2:30 P.M. and then speeded up considerably so as to arrive at the point 84 miles from A at 2:35 P.M. Evidently, a better approximation of the motion would be obtained by considering the average velocity during a smaller time interval, say from 2:30P.M. to 2:31 P.M. Indeed, it appears that the best procedure would be to take smaller and smaller time intervals near 2:30P.M. and study the average velocity in each time interval. This leads us into a limiting process similar to that discussed for tangent lines. In order to base our discussion on mathematical concepts, let us assume that the position of an object moving rectilinearly may be represented by a point P on a coordinate line /. We shall sometimes refer to the motion of the point P on l, or the motion of a particle on l whose position is specified by P. We further assume that the position of Pis known at every instant in a given interval of time. Iff(t) denotes the coordinate of Pat timet, then the function f determined in this way is called the position function for P. If we keep track of time by means of a clock, then for each t the point Pis f(t) units from the origin, as illustrated in Figure 3.3. Time Position of P 0 p 0 f(t) FIGURE 3.3 99. 90 3 The Derivative Definition (3.3) To define the velocity of Pat time a, we begin by investigating the average velocity in a (small) time interval near a. Thus we consider times a and a + h where h may be positive or negative but not zero. The corresponding positions of P on the /-axis are given by f(a + h) and f(a), as illustrated in Figure 3.4, and hence the amount ofchange in the position of Pis f(a + h) - f(a). The latter number may be positive, negative, or zero, depending on whether the position of Pat time a + his to the right, to the left, or the same as its position at time a. This number is not necessarily the distance traversed by P during the time interval [a, a + h] since, for example, P may have moved beyond the point corresponding to f(a + h) and then returned to that point at time a + h. Change in time Change in position of P 0 f(a) f(a +h) FIGURE 3.4 By (3.2), the average velocity of P during the time interval [a, a + h] is given by Average velocity= f(a + h~ - f(a). As in our previous discussion, the smaller h is numerically, the closer this quotient should approximate the velocity of P at time a. Accordingly, we define the velocity as the limit, as h approaches zero, of the average velocity, provided the limit exists. This limit is also called the instantaneous velocity of Pat time a. Summarizing, we have the following definition. If a point P moves on a coordinate line l such that its coordinate at time t is f(t), then the velocity v(a) of Pat time a is given by ( ) I. f(a + h) - f(a) v a = 1m h h-0 provided the limit exists. If f(t) is measured in centimeters and t in seconds, then the unit of velocity is centimeters per second, abbreviated em/sec. If f(t) is in miles and t in hours, then velocity is in miles per hour (mijhr). Other units of measure- ment may, of course, be used. 100. 3.1 Exercises Introduction 3.1 9/ We shall return to the velocity concept in Chapter 4, where it will be shown that if the velocity is positive in a given time interval, then the point is moving in the positive direction on l, whereas if the velocity is negative, the point is moving in the negative direction. Although these facts have not been proved we shall use them in the following example. Example 2 The position of a point P on a coordinate line l is given by f(t) = t2 - 6t, where f(t) is measured in feet and t in seconds. Find the velocity at time a. What is the velocity at t = 0? At t = 4? Determine time intervals in which (a) P moves in the positive direction on land (b) P moves in the negative direction. At what time is the velocity 0? Solution From the formula for f(t) we obtain f(a + h) - f(a) h [(a + h)2 - 6(a + h)] - [a2 - 6a] h a2 + 2ah + h2 - 6a - 6h - a2 + 6a h 2ah + h2 - 6h h = 2a + h- 6. Consequently, by Definition (3.3), the velocity v(a) at time a is v(a) = lim f(a + h) - f(a) = lim (2a + h - 6) = 2a - 6. h->0 h h->0 In particular, the velocity at t = 0 is v(O) = 2(0) - 6 = -6 ft/sec. At t = 4 it is v(4) = 2(4)- 6 = 2ft/sec. According to the remarks preceding this example, P moves to the left when the velocity is negative; that is, when 2a - 6 < 0, or a < 3. The particle moves to the right when 2a - 6 > 0 or a > 3. In terms of intervals, the motion is to the left in the time interval (- w, 3) and to the right in (3, w ). The velocity is zero when 2a - 6 = 0, that is, when a = 3 seconds. You have undoubtedly noted the similarity of the limit in Definition (3.3) to that used in the definition of tangent line in (3.1). Indeed, the two ex- pressions are identical! There are many different mathematical and physical applications which lead to precisely this same limit. Several of these will be discussed in the next chapter. In Exercises 1-4 find the slope of the tangent line at the point P(a,j(a)) on the graph of f. Sketch the graph and show the taTJgect lines at various points. 1 f(x) = 2- x3 3 f(x) = Jx + 1 2 f(x) = 3x- 5 4 f(x) = (1/x) - 1 101. 92 3 The Derivative In Exercises 5 and 6, the position of a point P moving on a coordinate line lis given by f(t), where tis measured in seconds and f(t) in centimeters. is the velocity of the projectile at t = 2, t = 3, and 1 = 4? At what time does it reach its maximum height? When does it strike the ground? What is its velocity at the moment of impact? (a) Find the average velocity of P in the following time intervals: [1, 1.2]; [1, 1.1]; [1, 1.01]; [1, 1.001]. 8 If a balloonist drops ballast (a sand bag) from a balloon 500 ft above the ground, then its distance above ground after 1seconds is 500 - 1612 ft. Find the velocity at 1 = 1, t = 2, and t = 3. With what velocity does the ballast strike the ground? (b) Find the velocity of Pat t = I. (c) Determine the time intervals in which P moves in the positive direction. (d) Determine the time intervals in which P moves in the negative direction. 9 If the position of an object moving on a coordinate line is given by a polynomial function of degree 1, prove tht the velocity is constant.5 f(t) = 4t2 + 3t 6 j(l) = 13 7 A projectile is fired directly upward from the ground with an initial velocity of 112 ftjsec, and its distance above the ground after 1 seconds is 1121 - 1612 ft. What 10 If the position function of a rectilinearly moving object is a constant function, prove that the velocity is 0 at all times. Describe the motion of the particle. 3.2 The Derivative ofa Function Definition (3.4) In the preceding section the same limiting process occurred in two different situations. The limit in Definitions (3.1) and (3.3) is one of the fundamental concepts of calculus. In the remainder of the chapter a number of rules pertaining to this concept will be developed. Let us begin by introducing the terminology and notation associated with this limit. Let .f be a function that is defined on an open interval containing a. The derivative offat a, writtenf'(a), is given by .f'(a) = lim .f(a + h) - .f(a) h~O h provided the limit exists. The symbol f'(a) is read "f prime of a." The terminology "f'(a) exists" will mean that the limit in Definition (3.4) exists. If.f'(a) exists we say that the function f is differentiable at a orf has a derivative at a. It is important to observe that iffis differentiable at a, then by Definition (3.1 ),f'(a) is the slope of the tangent line to the graph offat the point (a,f(a)). We say that a functionfis differentiable on an open interval (a, h) if it is differentiable at every number c in (a, b). In like manner, we refer to functions that are differentiable on intervals of the form (a, oo), (- oo, a), or (- oo, oo). For closed intervals we use the following convention, which is analogous to the definition of continuity on a closed interval given in (2.25). 102. Definition (3.5) Slope= lim f(a +h)- f(a) h~ o h Slope= lim f(b +h)- f(b) h-o- h FIGURE 3.5 a The Derivative ofa Function 3.2 93 A functionfisdifferentiable onaclosedinterval [a, h] ifit is differentiable on (a, b) and if the following limits exist: lim f(a + h) - f{a) and h-+0+ h 1. f(b + h) - f(b) Jm h . h-+0- The one-sided limits specified in Definition (3.5) are sometimes referred to as the right-hand and left-hand derivatives offat a and b, respectively. Note that for the right-hand derivative we have h--+ o+. Consequently h > 0 and a +.h > a. Thus we may think ofa + has approaching afrom the right. For the left-hand derivative, h --+ o- and in this case, h < 0 and b + h < b. Hence for the left-hand derivative we may regard b + h as approaching b from the left. Iff is defined on a closed interval [a, b] and is undefined outside of this interval, then the right-hand and left-hand derivatives allow us to define the slopes of the tangent lines at the points P(a, f(a)) and R(b, f(b)), re- spectiveiy, as illustrated in Figure 3.5. Thus, for the slope of the tangent line at P we take the limiting value of the slope mPQ of the secant line through P and Qas Qapproaches P from the right. For the tangent line at R, the point Qapproaches R from the left. Differentiabilityonanintervaloftheform [a, b), [a, oo),(a, b],or(- oo, b] is define