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This is a special edition of an established title widely used by colleges and universities throughout the world Pearson published this exclusive edition for the benefit of students outside the United States If you purchased this book within the United States, you should be aware that it has been imported without the approval of the Publisher or Author • clear, easy-to-understand examples reinforced by application to real-world problems; and • an intuitive and less formal approach to the introduction of new or difficult concepts Also available for purchase is MyLab Math, a digital suite featuring a variety of problems based on textbook content, for additional practice It provides automatic grading and immediate answer feedback to allow students to review their understanding and gives instructors comprehensive feedback to help them monitor students’ progress University Calculus Early Transcendentals Fourth Edition in SI Units Joel Hass Christopher Heil Przemyslaw Bogacki Fourth Edition in SI Units The fourth edition also features updated graphics and new types of homework exercises, helping students visualize mathematical concepts clearly and acquire different perspectives on each topic University Calculus • precise explanations, carefully crafted exercise sets, and detailed solutions; Early Transcendentals University Calculus: Early Transcendentals provides a modern and streamlined treatment of calculus that helps students develop mathematical maturity and proficiency In its fourth edition, this easy-to-read and conversational text continues to pave the path to mastering the subject through GLOBAL EDITION G LO B A L EDITION GLOBAL EDITION Maurice D Weir George B Thomas, Jr Hass • Heil • Bogacki Weir • Thomas, Jr Hass_04_1292317302_Final.indd 11/05/19 6:42 AM MyLab Math for University Calculus, 4e in SI Units (access code required) MyLab™ Math is the teaching and learning platform that empowers instructors to reach every student By combining trusted author content with digital tools and a flexible platform, MyLab Math for University Calculus, 4e in SI Units, personalizes the learning experience and improves results for each student Interactive Figures A full suite of Interactive Figures was added to illustrate key concepts and allow manipulation Designed in the freely available G eoGebra software, these figures can be used in lecture as well as by students independently Questions that Deepen U nderstanding MyLab Math includes a variety of question types designed to help students succeed in the course In Setup & Solve questions, students show how they set up a problem as well as the solution, better mirroring what is required on tests Additional Conceptual Questions were written by faculty at Cornell University to support deeper, theoretical understanding of the key concepts in calculus Learning Catalytics Now included in all MyLab Math courses, this student response tool uses smartphones, tablets, or laptops to engage students in more interactive tasks and thinking during lecture Learning Catalytics™ fosters student engagement and peer-to-peer learning with real-time analytics https://www.pearsonmylabandmastering.com/global/ This page intentionally left blank A01_THOM6233_05_SE_WALK.indd 1/13/17 6:50 PM UNIVERSITY CALCULUS EARLY TRANSCENDENTALS Fourth Edition in SI Units Joel Hass University of California, Davis Christopher Heil Georgia Institute of Technology Przemyslaw Bogacki Old Dominion University Maurice D Weir Naval Postgraduate School George B Thomas, Jr Massachusetts Institute of Technology SI conversion by José Luis Zuleta Estrugo École Polytechnique Fédérale de Lausanne Director, Portfolio Management: Deirdre Lynch Executive Editor: Jeff Weidenaar Editorial Assistant: Jonathan Krebs Content Producer: Rachel S Reeve Managing Producer: Scott Disanno Producer: Shannon Bushee Manager, Courseware QA: Mary Durnwald Manager, Content Development: Kristina Evans Product Marketing Manager: Emily Ockay Product 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Joel Hass, Christopher Heil, Przemyslaw Bogacki, Maurice Weir, and George B Thomas, Jr to be identified as the authors of this work have been asserted by them in accordance with the Copyright, Designs and Patents Act 1988 Authorized adaptation from the United States edition, entitled University Calculus: Early Transcendentals, ISBN 978-0-13-499554-0, by Joel Hass, Christopher Heil, Przemyslaw Bogacki, Maurice Weir, and George B Thomas, Jr., published by Pearson Education, Inc., © 2020, 2016, 2012 All rights reserved No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without either the prior written permission of the publisher or a license permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, Saffron House, 6–10 Kirby Street, London EC1N 8TS All trademarks used herein are the property of their respective owners The use of any trademark in this text does not vest in the author or publisher any trademark ownership rights in such trademarks, nor does the use of such trademarks imply any affiliation with or endorsement of this book by such owners For information regarding permissions, request forms, and the appropriate contacts within the Pearson Education Global Rights and Permissions department, please visit www.pearsoned.com/permissions This eBook is a standalone product and may or may not include all assets that were part of the print version It also does not provide access to other Pearson digital products like MyLab and Mastering The publisher reserves the right to remove any material in this eBook at any time British Library Cataloging-in-Publication Data A catalogue record for this book is available from the British Library ISBN 10: 1-292-31730-2 ISBN 13: 978-1-292-31730-4 eBook ISBN 13: 978-1-292-31729-8 A02_HASS7304_04_GE_FM.indd 27/05/2019 20:14 Contents Preface Functions 19 1.1 Functions and Their Graphs 19 1.2 Combining Functions; Shifting and Scaling Graphs 32 1.3 Trigonometric Functions 39 1.4 Graphing with Software 47 1.5 Exponential Functions 51 1.6 Inverse Functions and Logarithms 56 Also available: A.1 Real Numbers and the Real Line, A.3 Lines and Circles 2.1 2.2 2.3 2.4 2.5 2.6 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11 Limits and Continuity 69 Rates of Change and Tangent Lines to Curves 69 Limit of a Function and Limit Laws 76 The Precise Definition of a Limit 87 One-Sided Limits 96 Continuity 103 Limits Involving Infinity; Asymptotes of Graphs 115 Questions to Guide Your Review 128 Practice Exercises 129 Additional and Advanced Exercises 131 Also available: A.5 Proofs of Limit Theorems Derivatives 134 Tangent Lines and the Derivative at a Point 134 The Derivative as a Function 138 Differentiation Rules 147 The Derivative as a Rate of Change 157 Derivatives of Trigonometric Functions 166 The Chain Rule 172 Implicit Differentiation 180 Derivatives of Inverse Functions and Logarithms 185 Inverse Trigonometric Functions 195 Related Rates 202 Linearization and Differentials 210 Questions to Guide Your Review 221 Practice Exercises 222 Additional and Advanced Exercises 226 Contents 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 5.4 5.5 5.6 6.1 6.2 6.3 6.4 6.5 6.6 Applications of Derivatives 230 Extreme Values of Functions on Closed Intervals 230 The Mean Value Theorem 238 Monotonic Functions and the First Derivative Test 246 Concavity and Curve Sketching 251 Indeterminate Forms and L’Hôpital’s Rule 264 Applied Optimization 273 Newton’s Method 284 Antiderivatives 289 Questions to Guide Your Review 299 Practice Exercises 300 Additional and Advanced Exercises 304 Integrals 308 Area and Estimating with Finite Sums 308 Sigma Notation and Limits of Finite Sums 318 The Definite Integral 325 The Fundamental Theorem of Calculus 338 Indefinite Integrals and the Substitution Method 350 Definite Integral Substitutions and the Area Between Curves 357 Questions to Guide Your Review 367 Practice Exercises 368 Additional and Advanced Exercises 371 Applications of Definite Integrals 374 Volumes Using Cross-Sections 374 Volumes Using Cylindrical Shells 385 Arc Length 393 Areas of Surfaces of Revolution 399 Work 404 Moments and Centers of Mass 410 Questions to Guide Your Review 419 Practice Exercises 420 Additional and Advanced Exercises 421 Integrals and Transcendental Functions 423 7.1 The Logarithm Defined as an Integral 423 7.2 Exponential Change and Separable Differential Equations 433 7.3 Hyperbolic Functions 443 Questions to Guide Your Review 451 Practice Exercises 451 Additional and Advanced Exercises 452 Also available: B.1 Relative Rates of Growth Contents Techniques of Integration 454 8.1 Integration by Parts 455 8.2 Trigonometric Integrals 463 8.3 Trigonometric Substitutions 469 8.4 Integration of Rational Functions by Partial Fractions 474 8.5 Integral Tables and Computer Algebra Systems 481 8.6 Numerical Integration 487 8.7 Improper Integrals 496 Questions to Guide Your Review 507 Practice Exercises 508 Additional and Advanced Exercises 510 Also available: B.2 Probability Infinite Sequences and Series 513 9.1 Sequences 513 9.2 Infinite Series 526 9.3 The Integral Test 536 9.4 Comparison Tests 542 9.5 Absolute Convergence; The Ratio and Root Tests 547 9.6 Alternating Series and Conditional Convergence 554 9.7 Power Series 561 9.8 Taylor and Maclaurin Series 572 9.9 Convergence of Taylor Series 577 9.10 Applications of Taylor Series 583 Questions to Guide Your Review 592 Practice Exercises 593 Additional and Advanced Exercises 595 Also available: A.6 Commonly Occurring Limits 10 Parametric Equations and Polar Coordinates 598 10.1 Parametrizations of Plane Curves 598 10.2 Calculus with Parametric Curves 606 10.3 Polar Coordinates 616 10.4 Graphing Polar Coordinate Equations 620 10.5 Areas and Lengths in Polar Coordinates 624 Questions to Guide Your Review 629 Practice Exercises 629 Additional and Advanced Exercises 631 Also available: A.4 Conic Sections, B.3 Conics in Polar Coordinates Contents 11 Vectors and the Geometry of Space 632 11.1 Three-Dimensional Coordinate Systems 632 11.2 Vectors 637 11.3 The Dot Product 646 11.4 The Cross Product 654 11.5 Lines and Planes in Space 660 11.6 Cylinders and Quadric Surfaces 669 Questions to Guide Your Review 675 Practice Exercises 675 Additional and Advanced Exercises 677 Also available: A.9 The Distributive Law for Vector Cross Products 12 12.1 12.2 12.3 12.4 12.5 12.6 13 13.1 13.2 13.3 13.4 13.5 13.6 13.7 13.8 Vector-Valued Functions and Motion in Space 680 Curves in Space and Their Tangents 680 Integrals of Vector Functions; Projectile Motion 689 Arc Length in Space 696 Curvature and Normal Vectors of a Curve 700 Tangential and Normal Components of Acceleration 705 Velocity and Acceleration in Polar Coordinates 708 Questions to Guide Your Review 712 Practice Exercises 712 Additional and Advanced Exercises 714 Partial Derivatives 715 Functions of Several Variables 715 Limits and Continuity in Higher Dimensions 723 Partial Derivatives 732 The Chain Rule 744 Directional Derivatives and Gradient Vectors 754 Tangent Planes and Differentials 762 Extreme Values and Saddle Points 772 Lagrange Multipliers 781 Questions to Guide Your Review 791 Practice Exercises 791 Additional and Advanced Exercises 795 Also available: A.10 The Mixed Derivative Theorem and the Increment Theorem, B.4 Taylor’s Formula for Two Variables, B.5 Partial Derivatives with Constrained Variables Contents 14 Multiple Integrals 797 14.1 Double and Iterated Integrals over Rectangles 797 14.2 Double Integrals over General Regions 802 14.3 Area by Double Integration 811 14.4 Double Integrals in Polar Form 814 14.5 Triple Integrals in Rectangular Coordinates 821 14.6 Applications 831 14.7 Triple Integrals in Cylindrical and Spherical Coordinates 838 14.8 Substitutions in Multiple Integrals 850 Questions to Guide Your Review 859 Practice Exercises 860 Additional and Advanced Exercises 862 15 15.1 15.2 15.3 15.4 15.5 15.6 15.7 15.8 16 Integrals and Vector Fields 865 Line Integrals of Scalar Functions 865 Vector Fields and Line Integrals: Work, Circulation, and Flux 872 Path Independence, Conservative Fields, and Potential Functions 885 Green’s Theorem in the Plane 896 Surfaces and Area 908 Surface Integrals 918 Stokes’ Theorem 928 The Divergence Theorem and a Unified Theory 941 Questions to Guide Your Review 952 Practice Exercises 952 Additional and Advanced Exercises 955 First-Order Differential Equations 16-1 (Online) 16.1 Solutions, Slope Fields, and Euler’s Method 16-1 16.2 First-Order Linear Equations 16-9 16.3 Applications 16-15 16.4 Graphical Solutions of Autonomous Equations 16-21 16.5 Systems of Equations and Phase Planes 16-28 Questions to Guide Your Review 16-34 Practice Exercises 16-34 Additional and Advanced Exercises 16-36 17 Second-Order Differential Equations 17-1 (Online) 17.1 Second-Order Linear Equations 17-1 17.2 Nonhomogeneous Linear Equations 17-7 17.3 Applications 17-15 17.4 Euler Equations 17-22 17.5 Power-Series Solutions 17-24 2.2 Limit of a Function and Limit Laws 83 THEOREM 4—The Sandwich Theorem Suppose that g(x) … ƒ(x) … h(x) for all x in some open interval containing c, except possibly at x = c itself Suppose also that lim g(x) = lim h(x) = L xSc xSc Then lim ƒ(x) = L xSc y EXAMPLE 10 Given a function u that satisfies y = u(x) -1 The Sandwich Theorem is also called the Squeeze Theorem or the Pinching Theorem y=1+ x 2 y=1- x x 1 - x2 x2 … u(x) … + for all x ≠ 0, find limxS0 u(x), no matter how complicated u is Solution Since FIGURE 2.13 Any function u(x) whose graph lies in the region between y = + (x2 >2) and y = - (x2 >4) has limit as x S (Example 10) lim (1 - (x2 >4)) = lim (1 + (x2 >2)) = 1, and xS0 xS0 the Sandwich Theorem implies that limxS0 u(x) = (Figure 2.13). EXAMPLE 11 The Sandwich Theorem helps us establish several important limit rules: (a) lim sin u = uS0 (b) lim cos u = uS0 (c) For any function ƒ, lim ƒ(x) = implies lim ƒ(x) = xSc y Solution y = 0u0 y = sin u -p u p lim sin u = (b) From Section 1.3, … - cos u … u for all u (see Figure 2.14b) Since lim u = uS0 and lim = 0, we have limuS0 (1 - cos u) = so (a) uS0 lim - (1 - cos u) = - lim (1 - cos u) = - 0, uS0 y (a) In Section 1.3 we established that - u … sin u … u for all u (see Figure 2.14a) Since limuS0(- u ) = limuS0 u = 0, we have uS0 y = - 0u0 -1 xSc uS0 lim cos u = y = 0u0 Simplify uS0 u (c) Since - ƒ(x) … ƒ(x) … ƒ(x) and - ƒ(x) and ƒ(x) have limit as x S c, it follows that limxSc ƒ(x) = 0. FIGURE 2.14 The Sandwich Theorem Example 11 shows that the sine and cosine functions are equal to their limits at u = We have not yet established that for any c, lim sin u = sin c, and lim cos u = cos c confirms the limits in Example 11 These limit formulas hold, as will be shown in Section 2.5 -2 -1 y = - cos u (b) uSc uSc 84 Chapter 2 Limits and Continuity EXERCISES 2.2 Limits from Graphs e lim ƒ(x) exists at every point c in (1, 3) 1 For the function g(x) graphed here, find the following limits or explain why they not exist ƒ(1) = f a lim g(x) b. lim g(x) c. lim g(x) d. lim g(x) xS1 xS2 xS3 x S 2.5 xSc g ƒ(1) = -2 y = f (x) h ƒ(2) = i ƒ(2) = y y -1 x -1 y = g(x) -2 x Existence of Limits In Exercises and 6, explain why the limits not exist For the function ƒ(t) graphed here, find the following limits or explain why they not exist a lim ƒ(t) b. lim ƒ(t) c. lim ƒ(t) d. lim ƒ(t) t S -2 t S -1 tS0 t S -0.5 s s = f (t) -1 -2 t lim xS0 x 0x0 6. lim xS1 x - Suppose that a function ƒ(x) is defined for all real values of x except x = c Can anything be said about the existence of limxSc ƒ(x)? Give reasons for your answer Suppose that a function ƒ(x) is defined for all x in 3- 1, 14 Can anything be said about the existence of limxS0 ƒ(x)? Give reasons for your answer If limxS1 ƒ(x) = 5, must ƒ be defined at x = 1? If it is, must ƒ(1) = 5? Can we conclude anything about the values of ƒ at x = 1? Explain -1 Which of the following statements about the function y = ƒ(x) graphed here are true, and which are false? a lim ƒ(x) exists 10 If ƒ(1) = 5, must limxS1 ƒ(x) exist? If it does, then must limxS1 ƒ(x) = 5? Can we conclude anything about limxS1 ƒ(x)? Explain xS0 Calculating Limits xS0 Find the limits in Exercises 11–22 xS0 lim (-x2 + 5x - 2) 11 lim (x2 - 13) 12. xS1 13. lim 8(t - 5)(t - 7) 14. lim (x3 - 2x2 + 4x + 8) xS1 2x + 15. lim 16. lim (8 - 3s)(2s - 1) x S 11 - x s S 2>3 lim ƒ(x) = b c lim ƒ(x) = x S -3 d lim ƒ(x) = tS6 e lim ƒ(x) = f lim ƒ(x) exists at every point c in (- 1, 1) xSc g lim ƒ(x) does not exist xS1 h ƒ(0) = 17. y i ƒ(0) = 1 j ƒ(1) = -1 y + lim 4x(3x + 4)2 18. lim y S y + 5y + x S -1>2 y S -3 x -1 21. lim hS0 Which of the following statements about the function y = ƒ(x) graphed here are true, and which are false? a lim ƒ(x) does not exist xS2 lim ƒ(x) = b xS2 c lim ƒ(x) does not exist lim ƒ(x) exists at every point c in (- 1, 1) d zS4 23h + + 25h + - 22. lim h hS0 Limits of quotients Find the limits in Exercises 23–42 23 lim xS5 xSc x S -2 19. lim (5 - y)4>3 20. lim 2z2 - 10 y = f (x) k ƒ(1) = - xS1 xS2 x - x + 24. lim x S -3 x + 4x + x2 - 25 x2 + 3x - 10 x2 - 7x + 10 26. lim x - x + x S -5 xS2 25. lim t2 + t - t + 3t + 28. lim 2 tS1 t - t S -1 t - t - 27 lim 29 lim x S -2 5y + 8y -2x - 30. lim y S 3y - 16y x + 2x 2.2 Limit of a Function and Limit Laws 1 x-1 - x - + x + 32. lim xS1 x - xS0 x u4 - y3 - 33 lim 34. lim uS1 u - y S y - 16 1x - 4x - x2 35 lim 36. lim xS9 x - x S - 1x 31 lim 37 lim xS1 53 Suppose limxSc ƒ(x) = and limxSc g(x) = -2 Find lim 2ƒ(x)g(x) a lim ƒ(x)g(x) b. xSc 54 Suppose limxS4 ƒ(x) = and limxS4 g(x) = -3 Find lim xƒ(x) a lim (g(x) + 3) b. 2x2 + - x - 38. lim x + x S -1 2x + - xS4 2x + 12 - x + 40. lim x - x S -2 2x + - - 2x2 - - x 41 lim 42. lim x + x S -3 x S - 2x + Limits with trigonometric functions Find the limits in Exercises 43–50 xS2 lim sin x 43 lim (2 sin x - 1) 44. xS0 45 lim sec x 46. lim tan x xS0 xS4 g(x) c lim (g(x))2 d. lim xS4 x S ƒ(x) - 55 Suppose limxSb ƒ(x) = and limxSb g(x) = -3 Find lim ƒ(x) # g(x) a lim ( ƒ(x) + g(x)) b. xSb xSb c lim 4g(x) d. lim ƒ(x)>g(x) xSb xSb 56 Suppose that limxS -2 p(x) = 4, limxS -2 r(x) = 0, and limxS -2 s(x) = -3 Find xS0 xSc ƒ(x) c lim (ƒ(x) + 3g(x)) d. lim xSc x S c ƒ(x) - g(x) 39 lim xS0 a lim (p(x) + r(x) + s(x)) x S -2 + x + sin x 47 lim 48. lim (x2 - 1)(2 - cos x) cos x xS0 xS0 b lim p(x) # r(x) # s(x) 49 lim 2x + cos (x + p) 50. lim 27 + sec2 x c lim (-4p(x) + 5r(x))>s(x) x S -p 51 Suppose limxS0 ƒ(x) = and limxS0 g(x) = - Name the rules in Theorem that are used to accomplish steps (a), (b), and (c) of the following calculation lim (2ƒ(x) - g(x)) 2ƒ(x) - g(x) xS0 lim = (a) x S ( ƒ(x) + 7)2 lim (ƒ(x) + 7)2 xS0 (We assume the denominator is nonzero.) lim 2ƒ(x) - lim g(x) xS0 xS0 (b) = a lim (ƒ(x) + 7)b xS0 = lim ƒ(x) - lim g(x) xS0 xS0 a lim ƒ(x) + lim 7b xS0 = (c) lim 25h(x) - r(x)) lim 25h(x) xS1 (a) lim (p(x)(4 - r(x))) xS1 lim 5h(x) xS1 (b) xS1 xS1 2(5)(5) = (1)(4 - 2) 57 ƒ(x) = x2, x = 58 ƒ(x) = x2, x = -2 59 ƒ(x) = 3x - 4, x = 60 ƒ(x) = 1>x, x = -2 61 ƒ(x) = 1x, x = xS1 63 If 25 - 2x2 … ƒ(x) … 25 - x2 for - … x … 1, find limxS0 ƒ(x) 64 If - x2 … g(x) … cos x for all x, find limxS0 g(x) 65 a. It can be shown that the inequalities - x2 x sin x 6 - cos x lim x sin x ? - cos x Give reasons for your answer (c) a lim p(x)b a lim - lim r(x)b xS1 = xS1 25lim h(x) = occur frequently in calculus In Exercises 57–62, evaluate this limit for the given value of x and function ƒ xS0 a lim p(x)b a lim (4 - r(x))b xS1 ƒ(x + h) - ƒ(x) h hS0 lim hold for all values of x close to zero (except for x = 0) What, if anything, does this tell you about (We assume the denominator is nonzero.) = Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form Using the Sandwich Theorem (2)(1) - (- 5) = 64 (1 + 7)2 = Limits of Average Rates of Change 62 ƒ(x) = 23x + 1, x = xS0 52 Let limxS1 h(x) = 5, limxS1 p(x) = 1, and limxS1 r(x) = Name the rules in Theorem that are used to accomplish steps (a), (b), and (c) of the following calculation x S p(x)(4 x S -2 x S -2 xS0 Using Limit Rules 85 T b Graph y = - (x2 >6), y = (x sin x)>(2 - cos x), and y = together for -2 … x … Comment on the behavior of the graphs as x S 86 Chapter 2 Limits and Continuity 66 a Suppose that the inequalities x2 - cos x 1 6 24 x2 hold for values of x close to zero, except for x = itself (They do, as you will see in Section 9.9.) What, if anything, does this tell you about lim xS0 - cos x ? x2 Give reasons for your answer T b Graph the equations y = (1>2) - (x2 >24), y = (1 - cos x)>x2, and y = 1>2 together for -2 … x … Comment on the behavior of the graphs as x S Estimating Limits T You will find a graphing calculator useful for Exercises 67–76 b Support your conclusion in part (a) by graphing ƒ near c = -1 and using Zoom and Trace to estimate y-values on the graph as x S -1 c Find limxS -1 ƒ(x) algebraically 72 Let F(x) = (x2 + 3x + 2)>(2 - x ) a Make tables of values of F at values of x that approach c = -2 from above and below Then estimate limxS -2 F(x) b Support your conclusion in part (a) by graphing F near c = -2 and using Zoom and Trace to estimate y-values on the graph as x S -2 c Find limxS -2 F(x) algebraically 73 Let g(u) = (sin u)>u a Make a table of the values of g at values of u that approach u0 = from above and below Then estimate limuS0 g(u) b Support your conclusion in part (a) by graphing g near u0 = 67 Let ƒ(x) = (x2 - 9)>(x + 3) 74 Let G(t) = (1 - cos t)>t a Make a table of the values of ƒ at the points x = -3.1, - 3.01, -3.001, and so on as far as your calculator can go Then estimate limxS -3 ƒ(x) What estimate you arrive at if you evaluate ƒ at x = - 2.9, - 2.99, - 2.999, c instead? a Make tables of values of G at values of t that approach t0 = from above and below Then estimate limtS0 G(t) b Support your conclusions in part (a) by graphing ƒ near c = -3 and using Zoom and Trace to estimate y-values on the graph as x S - 75 Let ƒ(x) = x1>(1 - x) c Find limxS -3 ƒ(x) algebraically, as in Example 68 Let g(x) = (x2 - 2)>(x - 22) a Make a table of the values of g at the points x = 1.4, 1.41, 1.414, and so on through successive decimal approximations of 22 Estimate limxS 22 g(x) b Support your conclusion in part (a) by graphing G near t0 = a Make tables of values of ƒ at values of x that approach c = from above and below Does ƒ appear to have a limit as x S 1? If so, what is it? If not, why not? b Support your conclusions in part (a) by graphing ƒ near c = 76 Let ƒ(x) = (3x - 1)>x a Make tables of values of ƒ at values of x that approach c = from above and below Does ƒ appear to have a limit as x S 0? If so, what is it? If not, why not? b Support your conclusion in part (a) by graphing g near c = 22 and using Zoom and Trace to estimate y-values on the graph as x S 22 b Support your conclusions in part (a) by graphing ƒ near c = 69 Let G(x) = (x + 6)>(x2 + 4x - 12) Theory and Examples c Find limxS 22 g(x) algebraically a Make a table of the values of G at x = - 5.9, - 5.99, - 5.999, and so on Then estimate limxS -6 G(x) What estimate you arrive at if you evaluate G at x = - 6.1, - 6.01, -6.001, c instead? 77 If x4 … ƒ(x) … x2 for x in 3-1, 14 and x2 … ƒ(x) … x4 for x -1 and x 1, at what points c you automatically know limxSc ƒ(x)? What can you say about the value of the limit at these points? b Support your conclusions in part (a) by graphing G and using Zoom and Trace to estimate y-values on the graph as x S - 78 Suppose that g(x) … ƒ(x) … h(x) for all x ≠ and suppose that c Find limxS -6 G(x) algebraically Can we conclude anything about the values of ƒ, g, and h at x = 2? Could ƒ(2) = 0? Could limxS2 ƒ(x) = 0? Give reasons for your answers ƒ(x) - = 1, find lim ƒ(x) 79 If lim xS4 x - xS4 70 Let h(x) = (x2 - 2x - 3)>(x2 - 4x + 3) a Make a table of the values of h at x = 2.9, 2.99, 2.999, and so on Then estimate limxS3 h(x) What estimate you arrive at if you evaluate h at x = 3.1, 3.01, 3.001, c instead? b Support your conclusions in part (a) by graphing h near c = and using Zoom and Trace to estimate y-values on the graph as x S c Find limxS3 h(x) algebraically 71 Let ƒ(x) = (x2 - 1)>( x - 1) a Make tables of the values of ƒ at values of x that approach c = -1 from above and below Then estimate limxS -1 ƒ(x) lim g(x) = lim h(x) = - xS2 80 If lim x S -2 ƒ(x) = 1, find x2 xS2 ƒ(x) lim a lim ƒ(x) b. x S -2 x S -2 x ƒ(x) - = 3, find lim ƒ(x) 81 a If lim xS2 x - xS2 ƒ(x) - b If lim = 4, find lim ƒ(x) xS2 x - xS2 2.3 The Precise Definition of a Limit ƒ(x) = 1, find x S x2 82 If lim 85 lim ƒ(x) lim a lim ƒ(x) b. xS0 xS0 x x4 - 16 x - 86 lim x3 - x2 - 5x - (x + 1)2 87 lim 21 + x - T 83 a. Graph g(x) = x sin (1>x) to estimate limxS0 g(x), zooming in on the origin as necessary b. Confirm your estimate in part (a) with a proof T 84 a. Graph h(x) = x2 cos (1>x3) to estimate limxS0 h(x), zooming in on the origin as necessary xS2 x S -1 87 x xS0 88 lim xS3 x - 2x2 + - b Confirm your estimate in part (a) with a proof 89 lim - cos x x sin x COMPUTER EXPLORATIONS Graphical Estimates of Limits 90 lim 2x2 - cos x xS0 xS0 In Exercises 85–90, use a CAS to perform the following steps: a Plot the function near the point c being approached b From your plot guess the value of the limit 2.3 The Precise Definition of a Limit We now turn our attention to the precise definition of a limit The early history of calculus saw controversy about the validity of the basic concepts underlying the theory Apparent contradictions were argued over by both mathematicians and philosophers These controversies were resolved by the precise definition, which allows us to replace vague phrases like “gets arbitrarily close to” in the informal definition with specific conditions that can be applied to any particular example With a rigorous definition, we can avoid misunderstandings, prove the limit properties given in the preceding section, and establish many important limits To show that the limit of ƒ(x) as x S c equals the number L, we need to show that the gap between ƒ(x) and L can be made “as small as we choose” if x is kept “close enough” to c Let us see what this requires if we specify the size of the gap between ƒ(x) and L EXAMPLE 1 Consider the function y = 2x - near x = Intuitively it seems clear that y is close to when x is close to 4, so limxS4(2x - 1) = However, how close to x = does x have to be so that y = 2x - differs from by, say, less than units? y y = 2x - Upper bound: y=9 To satisfy this Lower bound: y=5 x Restrict to this FIGURE 2.15 Keeping x within unit of x = will keep y within units of y = (Example 1) Solution We are asked: For what values of x is y - 2? To find the answer we first express y - in terms of x: y - = (2x - 1) - = 2x - The question then becomes: What values of x satisfy the inequality 2x - 2? To find out, we solve the inequality: 2x - -2 -1 6 6 2x - 2x 10 x x - Removing absolute value gives two inequalities Add to each term Solve for x Solve for x - Keeping x within unit of x = will keep y within units of y = (Figure 2.15). In the previous example we determined how close x must be to a particular value c to ensure that the outputs ƒ(x) of some function lie within a prescribed interval about a limit value L To show that the limit of ƒ(x) as x S c actually equals L, we must be able to show that the gap between ƒ(x) and L can be made less than any prescribed error, no matter how 88 Chapter 2 Limits and Continuity small, by holding x close enough to c To describe arbitrary prescribed errors, we introduce two constants, d (delta) and e (epsilon) These Greek letters are traditionally used to represent small changes in a variable or a function d is the Greek letter delta e is the Greek letter epsilon Definition of Limit y L+ 10 f(x) f(x) lies in here L L− 10 for all x ≠ c in here d d x c−d c x c+d FIGURE 2.16 How should we define d so that keeping x within the interval (c - d, c + d) will keep ƒ(x) within the interval ¢L - 1 ,L + ≤? 10 10 Suppose we are watching the values of a function ƒ(x) as x approaches c (without taking on the value c itself) Certainly we want to be able to say that ƒ(x) stays within one-tenth of a unit from L as soon as x stays within some distance d of c (Figure 2.16) But that in itself is not enough, because as x continues on its course toward c, what is to prevent ƒ(x) from jumping around within the interval from L - (1>10) to L + (1>10) without tending toward L? We can be told that the error can be no more than 1>100 or 1>1000 or 1>100,000 Each time, we find a new d@interval about c so that keeping x within that interval satisfies the new error tolerance And each time the possibility exists that ƒ(x) might jump away from L at some later stage The figures on the next page illustrate the problem You can think of this as a quarrel between a skeptic and a scholar The skeptic presents e@challenges to show there is room for doubt that the limit exists The scholar counters every challenge with a d@interval around c which ensures that the function takes values within e of L How we stop this seemingly endless series of challenges and responses? We can so by proving that for every error tolerance e that the challenger can produce, we can present a matching distance d that keeps x “close enough” to c to keep ƒ(x) within that e@tolerance of L (Figure 2.17) This leads us to the precise definition of a limit DEFINITION Let ƒ(x) be defined on an open interval about c, except possibly at c itself We say that the limit of ƒ(x) as x approaches c is the number L, and write lim ƒ(x) = L, y xSc if, for every number e 0, there exists a corresponding number d such that ƒ(x) - L e whenever x - c d L+e L f(x) f(x) lies in here L-e for all x Z c in here d x c-d d x c c+d FIGURE 2.17 The relation of d and e in the definition of limit To visualize the definition, imagine machining a cylindrical shaft to a close tolerance The diameter of the shaft is determined by turning a dial to a setting measured by a variable x We try for diameter L, but since nothing is perfect we must be satisfied with a diameter ƒ(x) somewhere between L - e and L + e The number d is our control tolerance for the dial; it tells us how close our dial setting must be to the setting x = c in order to guarantee that the diameter ƒ(x) of the shaft will be accurate to within e of L As the tolerance for error becomes stricter, we may have to adjust d The value of d, how tight our control setting must be, depends on the value of e, the error tolerance The definition of limit extends to functions on more general domains It is only required that each open interval around c contain points in the domain of the function other than c See Additional and Advanced Exercises 39–43 for examples of limits for functions with complicated domains In the next section we will see how the definition of limit applies at points lying on the boundary of an interval Examples: Testing the Definition The formal definition of limit does not tell how to find the limit of a function, but it does enable us to verify that a conjectured limit value is correct The following examples show how the definition can be used to verify limit statements for specific functions However, the real purpose of the definition is not to calculations like this, but rather to prove general theorems so that the calculation of specific limits can be simplified, such as the theorems stated in the previous section 2.3 The Precise Definition of a Limit y L+ y = f (x) 10 L+ L- 10 x c c c + d1/10 c - d1/10 Response: x - c d1/10 (a number) The challenge: Make f(x) - L e = 10 L+ 100 L- L 100 x 100 L L100 L+ New challenge: Make f (x) - L e = y = f (x) 100 x y = f (x) L+ 1000 L L L1000 L1000 x c x c New challenge: e = 1000 Response: x - c d1/1000 y y y = f (x) L+ 100,000 L L 100,000 L- c New challenge: e = 100,000 y y = f (x) 100,000 c c + d1/100 c - d1/100 Response: x - c d1/100 y L+ 1000 L- x c y L+ y = f(x) y = f(x) L 10 y y = f(x) 10 L L- y y 89 x y = f(x) L+e L L-e 100,000 c x c Response: x - c d1/100,000 x New challenge: e = EXAMPLE 2 Show that lim (5x - 3) = xS1 Solution Set c = 1, ƒ(x) = 5x - 3, and L = in the definition of limit For any given e 0, we have to find a suitable d so that if x ≠ and x is within distance d of c = 1, that is, whenever x - d, it is true that ƒ(x) is within distance e of L = 2, so ƒ(x) - e 90 Chapter 2 Limits and Continuity y We find d by working backward from the e@inequality: y = 5x - (5x - 3) - = 5x - e 50x - 10 e x - e >5 2+e 2-e 1-e x 1+e Thus, we can take d = e >5 (Figure 2.18) If x - d = e >5, then (5x - 3) - = 5x - = x - 5(e >5) = e, which proves that limxS1 (5x - 3) = The value of d = e >5 is not the only value that will make x - d imply 5x - e Any smaller positive d will as well The definition does not ask for the “best” positive d, just one that will work. EXAMPLE 3 Prove the following results presented graphically in Section 2.2 -3 (a) lim x = c NOT TO SCALE xSc FIGURE 2.18 If ƒ(x) = 5x - 3, then (b) lim k = k (k constant) x - e >5 guarantees that ƒ(x) - e (Example 2) xSc Solution (a) Let e be given We must find d such that y 0x - c0 e whenever 0k - k0 e whenever x - c d The implication will hold if d equals e or any smaller positive number (Figure 2.19) This proves that limxSc x = c (b) Let e be given We must find d such that y=x c+e c+d c c-d x - c d Since k - k = 0, we can use any positive number for d and the implication will hold (Figure 2.20) This proves that limxSc k = k. c-e c-d c c+d x FIGURE 2.19 For the function ƒ(x) = x, we find that x - c d will guarantee ƒ(x) - c e whenever d … e (Example 3a) Finding Deltas Algebraically for Given Epsilons In Examples and 3, the interval of values about c for which ƒ(x) - L was less than e was symmetric about c and we could take d to be half the length of that interval When the interval around c on which we have ƒ(x) - L e is not symmetric about c, we can take d to be the distance from c to the interval’s nearer endpoint EXAMPLE 4 For the limit limxS5 2x - = 2, find a d that works for e = That is, find a d such that y y=k k+e k k-e 0 2x - - whenever x - d Solution We organize the search into two steps Solve the inequality 2x - - to find an interval containing x = on which the inequality holds for all x ≠ c-d c c+d x FIGURE 2.20 For the function ƒ(x) = k, we find that ƒ(x) - k e for any positive d (Example 3b) 2x - - -1 2x - - 1 2x - 6 x - x 10 2.3 The Precise Definition of a Limit ( 3 ) x 10 FIGURE 2.21 An open interval of radius about x = will lie inside the open interval (2, 10) y 2x - - whenever x - 3. The process of finding a d such that ƒ(x) - L e 0x - c0 d whenever can be accomplished in two steps The inequality holds for all x in the open interval (2, 10), so it holds for all x ≠ in this interval as well Find a value of d to place the centered interval - d x + d (centered at x = 5) inside the interval (2, 10) The distance from to the nearer endpoint of (2, 10) is (Figure 2.21) If we take d = or any smaller positive number, then the inequality x - d will automatically place x between and 10 and imply that 2x - - (Figure 2.22): How to Find Algebraically a D for a Given ƒ, L, c, and E + y = "x - 3 10 91 x NOT TO SCALE FIGURE 2.22 The function and intervals in Example Solve the inequality ƒ(x) - L e to find an open interval (a, b) containing c on which the inequality holds for all x ≠ c Note that we not require the inequality to hold at x = c It may hold there or it may not, but the value of ƒ at x = c does not influence the existence of a limit Find a value of d that places the open interval (c - d, c + d) centered at c inside the interval (a, b) The inequality ƒ(x) - L e will hold for all x ≠ c in this d@interval EXAMPLE 5 Prove that limxS2 ƒ(x) = if ƒ(x) = b x2, 1, x≠2 x = Solution Our task is to show that given e 0, there exists a d such that ƒ(x) - e y = x2 4+e x2 - e - e x2 - e - e x2 + e 24 - e x 24 + e 24 - e x 24 + e (2, 4) 4-e (2, 1) "4 - e x "4 + e FIGURE 2.23 An interval containing x = so that the function in Example satisfies ƒ(x) - e x - d Solve the inequality ƒ(x) - e to find an open interval containing x = on which the inequality holds for all x ≠ For x ≠ c = 2, we have ƒ(x) = x2, and the inequality to solve is x2 - e: y whenever Assumes e 4; see below An open interval about x = that solves the inequality The inequality ƒ(x) - e holds for all x ≠ in the open interval ( 24 - e, 24 + e) (Figure 2.23) Find a value of d that places the centered interval (2 - d, + d) inside the interval 24 - e, 24 + e Take d to be the distance from x = to the nearer endpoint of 24 - e, 24 + e In other words, take d = - 24 - e, 24 + e - , the minimum (the smaller) of the two numbers - 24 - e and 24 + e - If d has this or any 92 Chapter 2 Limits and Continuity smaller positive value, the inequality x - d will automatically place x between 24 - e and 24 + e to make ƒ(x) - e For all x, ƒ(x) - e whenever x - d This completes the proof for e If e Ú 4, then we take d to be the distance from x = to the nearer endpoint of the interval 0, 24 + e In other words, take d = 2, 24 + e - (See Figure 2.23.) Using the Definition to Prove Theorems We not usually rely on the formal definition of limit to verify specific limits such as those in the preceding examples Rather, we appeal to general theorems about limits, in particular the theorems of Section 2.2 The definition is used to prove these theorems (Appendix A.5) As an example, we prove part of Theorem 1, the Sum Rule EXAMPLE 6 Given that limxSc ƒ(x) = L and limxSc g(x) = M, prove that lim (ƒ(x) + g(x)) = L + M xSc Solution Let e be given We want to find a positive number d such that ƒ(x) + g(x) - (L + M) e x - c d whenever Regrouping terms, we get ƒ(x) + g(x) - (L + M) = (ƒ(x) - L) + (g(x) - M) … ƒ(x) - L + g(x) - M Since limxSc ƒ(x) = L, there exists a number d1 such that ƒ(x) - L e >2 whenever x - c d1 Triangle Inequality: 0a + b0 … 0a0 + 0b0 Can find d1 since lim ƒ(x) = L xSc Similarly, since limxSc g(x) = M, there exists a number d2 such that g(x) - M e >2 whenever x - c d2 Can find d2 since lim g(x) = M xSc Let d = d1, d2 , the smaller of d1 and d2 If x - c d then x - c d1, so ƒ(x) - L e >2, and x - c d2, so g(x) - M e >2 Therefore, ƒ(x) + g(x) - (L + M) e + e = e 2 This shows that limxSc(ƒ(x) + g(x)) = L + M. EXERCISES 2.3 Centering Intervals About a Point In Exercises 1–6, sketch the interval (a, b) on the x-axis with the point c inside Then find a value of d such that a x b whenever x - c d a = 1, b = 7, c = a = 1, b = 7, c = a = -7>2, b = -1>2, c = -3 a = -7>2, b = -1>2, c = -3>2 a = 4>9, b = 4>7, c = 1>2 a = 2.7591, b = 3.2391, c = 2.3 The Precise Definition of a Limit 13 Finding Deltas Graphically In Exercises 7–14, use the graphs to find a d such that ƒ(x) - L e whenever x - c d y y = 2x - 6.2 5.8 f (x) = 2x - c=5 L=6 e = 0.2 y= y= - 3x+3 x 4.9 f (x) = "-x c = -1 L=2 e = 0.5 y f (x) = - x + c = -3 L = 7.5 e = 0.15 14 y y f(x) = 1x c= L=2 e = 0.01 2.01 "-x 93 2.5 1.99 7.65 7.5 7.35 y = 1x 1.5 5.1 NOT TO SCALE -3.1 -3 -2.9 x -1 16 16 25 x 1 2.01 1.99 NOT TO SCALE NOT TO SCALE y 4 10 f(x) = " x c=1 L=1 e = y = "x 16 Each of Exercises 15–30 gives a function ƒ(x) and numbers L, c, and e In each case, find the largest open interval about c on which the inequality ƒ(x) - L e holds Then give a value for d such that for all x satisfying x - c d, the inequality ƒ(x) - L e holds f (x) = " x + c=3 L=4 e = 0.2 x 25 16 Finding Deltas Algebraically y y = "x + 4.2 3.8 15 ƒ(x) = x + 1, 16 ƒ(x) = 17 ƒ(x) = 18 ƒ(x) = -1 19 ƒ(x) = x 2.61 3.41 20 ƒ(x) = 21 ƒ(x) = NOT TO SCALE 22 ƒ(x) = 11 12 y y = x2 y = - x2 23 ƒ(x) = y f (x) = - x c = -1 L=3 e = 0.25 f(x) = x c=2 L=4 e=1 24 ƒ(x) = 25 ƒ(x) = 26 ƒ(x) = 3.25 27 ƒ(x) = L = 5, e = 0.01 2x - 2, L = -6, c = - 2, e = 0.02 2x + 1, L = 1, c = 0, e = 0.1 1x, L = 1>2, c = 1>4, e = 0.1 219 - x, L = 3, c = 10, e = 2x - 7, L = 4, c = 23, e = 1>x, L = 1>4, c = 4, e = 0.05 x 2, L = 3, c = 23, e = 0.1 x 2, L = 4, c = -2, e = 0.5 1>x, L = - 1, c = - 1, e = 0.1 x2 - 5, L = 11, c = 4, e = 120>x, L = 5, c = 24, e = mx, m 0, L = 2m, c = 2, e = 0.03 mx, m 0, L = 3m, c = 3, e = c 28 ƒ(x) = 2.75 29 ƒ(x) = mx + b, m 0, c = 1>2, e = c c = 4, L = (m>2) + b, 30 ƒ(x) = mx + b, m 0, L = m + b, c = 1, e = 0.05 x "3 Using the Formal Definition x Each of Exercises 31–36 gives a function ƒ(x), a point c, and a positive number e Find L = lim ƒ(x) Then find a number d such that "5 xSc NOT TO SCALE - " -1 " 2 NOT TO SCALE x ƒ(x) - L e whenever x - c d 31 ƒ(x) = - 2x, 32 ƒ(x) = - 3x - 2, 33 ƒ(x) = x2 - , x - c = 3, c = -1, c = 2, e = 0.02 e = 0.03 e = 0.05 94 Chapter 2 Limits and Continuity 34 ƒ(x) = x2 + 6x + , x + c = - 5, 35 ƒ(x) = 21 - 5x, 36 ƒ(x) = 4>x, c = - 3, c = 2, Theory and Examples e = 0.05 51 Define what it means to say that lim g(x) = k xS0 e = 0.5 52 Prove that lim ƒ(x) = L if and only if lim ƒ(h + c) = L xSc e = 0.4 Prove the limit statements in Exercises 37–50 The number L is the limit of ƒ(x) as x approaches c if ƒ(x) gets closer to L as x approaches c lim (3x - 7) = 37 lim (9 - x) = 5 38 xS4 xS3 39 lim 2x - = 2 40 lim 24 - x = xS9 Explain why the function in your example does not have the given value of L as a limit as x S c xS0 x 2, 41 lim ƒ(x) = if ƒ(x) = e xS1 2, 42 lim ƒ(x) = if ƒ(x) = e x S -2 x≠1 x = 54 Another wrong statement about limits Show by example that the following statement is wrong x ≠ -2 x = -2 x, 1, The number L is the limit of ƒ(x) as x approaches c if, given any e 0, there exists a value of x for which ƒ(x) - L e 1 43 lim x = 1 44 lim = xS1 x S 23 x Explain why the function in your example does not have the given value of L as a limit as x S c x2 - x2 - 45 lim lim = = - 6 46 xS1 x - x S -3 x + T 55 Grinding engine cylinders Before contracting to grind engine cylinders to a cross-sectional area of 60 cm2, you need to know how much deviation from the ideal cylinder diameter of c = 8.7404 cm you can allow and still have the area come within 0.1 cm2 of the required 60 cm2 To find out, you let A = p(x>2)2 and look for the interval in which you must hold x to make A - 60 … 0.1 What interval you find? - 2x, x 47 lim ƒ(x) = if ƒ(x) = e xS1 6x - 4, x Ú 2x, x 48 lim ƒ(x) = if ƒ(x) = e xS0 x>2, x Ú 49 lim x sin x = xS0 56 Manufacturing electrical resistors Ohm’s law for electrical circuits like the one VI R shown in the accompanying + figure states that V = RI In this equation, V is a constant voltage, I is the current in amperes, and R is the resistance in ohms Your firm has been asked to supply the resistors for a circuit in which V will be 120 volts and I is to be { 0.1 amp In what interval does R have to lie for I to be within 0.1 amp of the value I0 = 5? y - 2p y = x sin 1x 2p -p hS0 53 A wrong statement about limits Show by example that the following statement is wrong x p When Is a Number L Not the Limit of f (x) As x u c? Showing L is not a limit We can prove that limxSc ƒ(x) ≠ L by providing an e such that no possible d satisfies the condition ƒ(x) - L e whenever x - c d We accomplish this for our candidate e by showing that for each d 50 lim x2 sin x = there exists a value of x such that xS0 0x - c0 d y and y y = x2 ƒ(x) - L Ú e y = f (x) L+e y = x sin 1x -1 -p p L x L-e f (x) -1 y = -x c- d c c+ d A value of x for which x - c d and f (x) - L ≥ e x 2.3 The Precise Definition of a Limit 57 Let ƒ(x) = b y x x x, x + 1, 95 y 4.8 y=x+1 y = f (x) y = f (x) x x 60 a. For the function graphed here, show that limxS -1 g(x) ≠ b. Does limxS -1 g(x) appear to exist? If so, what is the value of the limit? If not, why not? y=x y a. Let e = 1>2 Show that no possible d satisfies the following condition: ƒ(x) - 1>2 whenever x - d That is, show that for each d 0, there is a value of x such that 0x - 10 d and This will show that limxS1 ƒ(x) ≠ y = g(x) ƒ(x) - Ú 1>2 b. Show that limxS1 ƒ(x) ≠ -1 x c. Show that limxS1 ƒ(x) ≠ 1.5 x 2, 58 Let h(x) = c 3, 2, x x = x COMPUTER EXPLORATIONS In Exercises 61–66, you will further explore finding deltas graphically Use a CAS to perform the following steps: a Plot the function y = ƒ(x) near the point c being approached y b Guess the value of the limit L and then evaluate the limit symbolically to see if you guessed correctly y = h(x) c Using the value e = 0.2, graph the banding lines y1 = L - e and y2 = L + e together with the function ƒ near c d From your graph in part (c), estimate a d such that y=2 y = x2 ƒ(x) - L e whenever x - c d x Show that Test your estimate by plotting ƒ, y1, and y2 over the interval x - c d For your viewing window use c - 2d … x … c + 2d and L - 2e … y … L + 2e If any function values lie outside the interval 3L - e, L + e 4, your choice of d was too large Try again with a smaller estimate lim h(x) ≠ a. e Repeat parts (c) and (d) successively for e = 0.1, 0.05, and 0.001 b. lim h(x) ≠ 61 ƒ(x) = xS2 xS2 c. lim h(x) ≠ xS2 59 For the function graphed here, explain why lim ƒ(x) ≠ a. xS3 b. lim ƒ(x) ≠ 4.8 xS3 c. lim ƒ(x) ≠ xS3 5x3 + 9x2 x4 - 81 , c = 3 62 ƒ(x) = , c = x - 2x + 3x2 x(1 - cos x) sin 2x , c = 0 64 , c = 63 ƒ(x) = ƒ(x) = 3x x - sin x 65 ƒ(x) = 2x - 66 ƒ(x) = 3x2 - (7x + 1) 1x + , c = x - x - , c = 96 Chapter 2 Limits and Continuity 2.4 One-Sided Limits In this section we extend the limit concept to one-sided limits, which are limits as x approaches the number c from the left-hand side (where x c ) or the right-hand side (where x c) only These allow us to describe functions that have different limits at a point, depending on whether we approach the point from the left or from the right Onesided limits also allow us to say what it means for a function to have a limit at an endpoint of an interval y Approaching a Limit from One Side y= x 0x0 x -1 FIGURE 2.24 Different right-hand and left-hand limits at the origin Suppose a function ƒ is defined on an interval that extends to both sides of a number c In order for ƒ to have a limit L as x approaches c, the values of ƒ(x) must approach the value L as x approaches c from either side Because of this, we sometimes say that the limit is two-sided If ƒ fails to have a two-sided limit at c, it may still have a one-sided limit, that is, a limit if the approach is only from one side If the approach is from the right, the limit is a right-hand limit or limit from the right Similarly, a left-hand limit is also called a limit from the left The function ƒ(x) = x> x (Figure 2.24) has limit as x approaches from the right, and limit -1 as x approaches from the left Since these one-sided limit values are not the same, there is no single number that ƒ(x) approaches as x approaches So ƒ(x) does not have a (two-sided) limit at Intuitively, if we consider only the values of ƒ(x) on an interval (c, b), where c b, and the values of ƒ(x) become arbitrarily close to L as x approaches c from within that interval, then ƒ has right-hand limit L at c In this case we write lim ƒ(x) = L x S c+ The notation “x S c+ ” means that we consider only values of ƒ(x) for x greater than c We don’t consider values of ƒ(x) for x … c Similarly, if ƒ(x) is defined on an interval (a, c), where a c, and ƒ(x) approaches arbitrarily close to M as x approaches c from within that interval, then ƒ has left-hand limit M at c We write lim ƒ(x) = M x S c- The symbol “x S c-” means that we consider the values of ƒ only at x-values less than c These informal definitions of one-sided limits are illustrated in Figure 2.25 For the function ƒ(x) = x> x in Figure 2.24 we have lim ƒ(x) = and x S 0+ lim ƒ(x) = -1 x S 0- y y f (x) L c x (a) lim + f (x) = L x: c M f (x) x x (b) c lim _ f (x) = M x: c FIGURE 2.25 (a) Right-hand limit as x approaches c (b) Left-hand limit as x approaches c x 2.4 One-Sided Limits 97 One-sided limits have all the properties listed in Theorem in Section 2.2 The right-hand limit of the sum of two functions is the sum of their right-hand limits, and so on The theorems for limits of polynomials and rational functions hold with one-sided limits, as does the Sandwich Theorem One-sided limits are related to limits at interior points in the following way THEOREM Suppose that a function ƒ is defined on an open interval containing c, except perhaps at c itself Then ƒ(x) has a limit as x approaches c if and only if it has both a limit from the left at c and a limit from the right at c, and these one-sided limits are equal: lim ƒ(x) = L xSc lim ƒ(x) = L and x S c- lim ƒ(x) = L x S c+ Theorem applies at interior points of a function’s domain At a boundary point of an interval in its domain, a function has a limit when it has an appropriate one-sided limit Limits at Endpoints of an Interval • If ƒ is defined on an open interval (b, c) to the left of c and not defined on an open interval (c, d) to the right of c, then lim ƒ(x) = lim- ƒ(x) xSc xSc • If ƒ is defined on an open interval (c, d) to the right of c and not defined on an open interval (b, c) to the left of c, then lim ƒ(x) = lim+ ƒ(x) xSc xSc (The definition of a limit on an arbitrary domain is discussed in Additional and Advanced Exercises 39–42.) EXAMPLE 1 For the function graphed in Figure 2.26, At x = 0: y y = f (x) ƒ is not defined to the left of x = ƒ has a right-hand limit at x = ƒ has a limit at domain endpoint x = At x = 1: limxS0- ƒ(x) does not exist, limxS0+ ƒ(x) = 1, limxS0 ƒ(x) = 1 x FIGURE 2.26 Graph of the function in Example y p y = arcsec x p -2 -1 x FIGURE 2.27 The arcsec function has limits at x = {1 Even though ƒ(1) = limxS1- ƒ(x) = 0, limxS1+ ƒ(x) = 1, Right- and left-hand limits are not equal limxS1 ƒ(x) does not exist At x = 2: limxS2- ƒ(x) = 1, limxS2+ ƒ(x) = 1, Even though ƒ(2) = limxS2 ƒ(x) = At x = 3: limxS3- ƒ(x) = limxS3+ ƒ(x) = limxS3 ƒ(x) = ƒ(3) = Even though ƒ(4) ≠ At x = 4: limxS4- ƒ(x) = 1, ƒ is not defined to the right of x = limxS4+ ƒ(x) does not exist, ƒ has a limit at domain endpoint x = limxS4 ƒ(x) = At every other point c in 0, 44 , ƒ(x) has limit ƒ(c). EXAMPLE 2 The domain of the function arcsec x is a union of the intervals (-q, -14 and 1, q), as shown in Figure 2.27 For the boundary points of these intervals, arcsec x has a limit from the left at x = - At x = -1: limxS -1- arcsec x = p, arcsec x is not defined on (- 1, 1) limxS -1+ arcsec x does not exist, arcsec x has a limit at x = - limxS -1 arcsec x = p ... Early Transcendentals, ISBN 978-0-13-499554-0, by Joel Hass, Christopher Heil, Przemyslaw Bogacki, Maurice Weir, and George B Thomas, Jr., published by Pearson Education, Inc., © 2020, 2016, 2012... www.pearsonglobaleditions.com © 2020 by Pearson Education, Inc Published by Pearson Education, Inc or its affiliates The rights of Joel Hass, Christopher Heil, Przemyslaw Bogacki, Maurice Weir, and George B Thomas, Jr.. . 1/13/17 6:50 PM UNIVERSITY CALCULUS EARLY TRANSCENDENTALS Fourth Edition in SI Units Joel Hass University of California, Davis Christopher Heil Georgia Institute of Technology Przemyslaw Bogacki Old