SOLUTION MANUAL

SOLUTION MANUAL 608070 _ISM_ThomasCalc_WeirHass_ttl.qxd:harsh_569709_ttl 9/3/09 3:11 PM Page 1 INSTRUCTOR’S SOLUTIONS MANUAL SINGLE VARIABLE Collin County Community College WILLIAM ARDIS THOMAS’ CALCULUS TWELFTH EDITION BASED ON THE ORIGINAL WORK BY George B. Thomas, Jr. Massachusetts Institute of Technology AS REVISED BY Maurice D. Weir Naval Postgraduate School Joel Hass University of California, Davis 608070 _ISM_ThomasCalc_WeirHass_ttl.qxd:harsh_569709_ttl 9/3/09 3:11 PM Page 2 This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials. The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. Reproduced by Pearson Addison-Wesley from electronic files supplied by the author. Copyright © 2010, 2005, 2001 Pearson Education, Inc. Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116. 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 the prior written permission of the publisher. Printed in the United States of America. ISBN-13: 978-0-321-60807-9 ISBN-10: 0-321-60807-0 1 2 3 4 5 6 BB 12 11 10 09 PREFACE TO THE INSTRUCTOR This Instructor's Solutions Manual contains the solutions to every exercise in the 12th Edition of THOMAS' CALCULUS by Maurice Weir and Joel Hass, including the Computer Algebra System (CAS) exercises. The corresponding Student's Solutions Manual omits the solutions to the even-numbered exercises as well as the solutions to the CAS exercises (because the CAS command templates would give them all away). In addition to including the solutions to all of the new exercises in this edition of Thomas, we have carefully revised or rewritten every solution which appeared in previous solutions manuals to ensure that each solution ì conforms exactly to the methods, procedures and steps presented in the text ì is mathematically correct ì includes all of the steps necessary so a typical calculus student can follow the logical argument and algebra ì includes a graph or figure whenever called for by the exercise, or if needed to help with the explanation ì is formatted in an appropriate style to aid in its understanding Every CAS exercise is solved in both the MAPLE and MATHEMATICA computer algebra systems. A template showing an example of the CAS commands needed to execute the solution is provided for each exercise type. Similar exercises within the text grouping require a change only in the input function or other numerical input parameters associated with the problem (such as the interval endpoints or the number of iterations). For more information about other resources available with Thomas' Calculus, visit http://pearsonhighered.com. TABLE OF CONTENTS 1 Functions 1 1.1 1.2 1.3 1.4 Functions and Their Graphs 1 Combining Functions; Shifting and Scaling Graphs 8 Trigonometric Functions 19 Graphing with Calculators and Computers 26 Practice Exercises 30 Additional and Advanced Exercises 38 2 Limits and Continuity 43 2.1 2.2 2.3 2.4 2.5 2.6 Rates of Change and Tangents to Curves 43 Limit of a Function and Limit Laws 46 The Precise Definition of a Limit 55 One-Sided Limits 63 Continuity 67 Limits Involving Infinity; Asymptotes of Graphs 73 Practice Exercises 82 Additional and Advanced Exercises 86 3 Differentiation 93 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 Tangents and the Derivative at a Point 93 The Derivative as a Function 99 Differentiation Rules 109 The Derivative as a Rate of Change 114 Derivatives of Trigonometric Functions 120 The Chain Rule 127 Implicit Differentiation 135 Related Rates 142 Linearizations and Differentials 146 Practice Exercises 151 Additional and Advanced Exercises 162 4 Applications of Derivatives 167 4.1 4.2 4.3 4.4 4.5 4.6 4.7 Extreme Values of Functions 167 The Mean Value Theorem 179 Monotonic Functions and the First Derivative Test 188 Concavity and Curve Sketching 196 Applied Optimization 216 Newton's Method 229 Antiderivatives 233 Practice Exercises 239 Additional and Advanced Exercises 251 5 Integration 257 5.1 5.2 5.3 5.4 5.5 5.6 Area and Estimating with Finite Sums 257 Sigma Notation and Limits of Finite Sums 262 The Definite Integral 268 The Fundamental Theorem of Calculus 282 Indefinite Integrals and the Substitution Rule 290 Substitution and Area Between Curves 296 Practice Exercises 310 Additional and Advanced Exercises 320 6 Applications of Definite Integrals 327 6.1 6.2 6.3 6.4 6.5 6.6 Volumes Using Cross-Sections 327 Volumes Using Cylindrical Shells 337 Arc Lengths 347 Areas of Surfaces of Revolution 353 Work and Fluid Forces 358 Moments and Centers of Mass 365 Practice Exercises 376 Additional and Advanced Exercises 384 7 Transcendental Functions 389 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 Inverse Functions and Their Derivatives 389 Natural Logarithms 396 Exponential Functions 403 Exponential Change and Separable Differential Equations 414 ^ Indeterminate Forms and L'Hopital's Rule 418 Inverse Trigonometric Functions 425 Hyperbolic Functions 436 Relative Rates of Growth 443 Practice Exercises 447 Additional and Advanced Exercises 458 8 Techniques of Integration 461 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Integration by Parts 461 Trigonometric Integrals 471 Trigonometric Substitutions 478 Integration of Rational Functions by Partial Fractions 484 Integral Tables and Computer Algebra Systems 491 Numerical Integration 502 Improper Integrals 510 Practice Exercises 518 Additional and Advanced Exercises 528 9 First-Order Differential Equations 537 9.1 9.2 9.3 9.4 9.5 Solutions, Slope Fields and Euler's Method 537 First-Order Linear Equations 543 Applications 546 Graphical Solutions of Autonomous Equations 549 Systems of Equations and Phase Planes 557 Practice Exercises 562 Additional and Advanced Exercises 567 10 Infinite Sequences and Series 569 10.1 Sequences 569 10.2 Infinite Series 577 10.3 The Integral Test 583 10.4 Comparison Tests 590 10.5 The Ratio and Root Tests 597 10.6 Alternating Series, Absolute and Conditional Convergence 602 10.7 Power Series 608 10.8 Taylor and Maclaurin Series 617 10.9 Convergence of Taylor Series 621 10.10 The Binomial Series and Applications of Taylor Series 627 Practice Exercises 634 Additional and Advanced Exercises 642 TABLE OF CONTENTS 10 Infinite Sequences and Series 569 10.1 Sequences 569 10.2 Infinite Series 577 10.3 The Integral Test 583 10.4 Comparison Tests 590 10.5 The Ratio and Root Tests 597 10.6 Alternating Series, Absolute and Conditional Convergence 602 10.7 Power Series 608 10.8 Taylor and Maclaurin Series 617 10.9 Convergence of Taylor Series 621 10.10 The Binomial Series and Applications of Taylor Series 627 Practice Exercises 634 Additional and Advanced Exercises 642 11 Parametric Equations and Polar Coordinates 647 11.1 11.2 11.3 11.4 11.5 11.6 11.7 Parametrizations of Plane Curves 647 Calculus with Parametric Curves 654 Polar Coordinates 662 Graphing in Polar Coordinates 667 Areas and Lengths in Polar Coordinates 674 Conic Sections 679 Conics in Polar Coordinates 689 Practice Exercises 699 Additional and Advanced Exercises 709 12 Vectors and the Geometry of Space 715 12.1 12.2 12.3 12.4 12.5 12.6 Three-Dimensional Coordinate Systems 715 Vectors 718 The Dot Product 723 The Cross Product 728 Lines and Planes in Space 734 Cylinders and Quadric Surfaces 741 Practice Exercises 746 Additional Exercises 754 13 Vector-Valued Functions and Motion in Space 759 13.1 13.2 13.3 13.4 13.5 13.6 Curves in Space and Their Tangents 759 Integrals of Vector Functions; Projectile Motion 764 Arc Length in Space 770 Curvature and Normal Vectors of a Curve 773 Tangential and Normal Components of Acceleration 778 Velocity and Acceleration in Polar Coordinates 784 Practice Exercises 785 Additional Exercises 791 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 14 Partial Derivatives 795 14.1 14.2 14.3 14.4 14.5 14.6 14.7 14.8 14.9 14.10 Functions of Several Variables 795 Limits and Continuity in Higher Dimensions 804 Partial Derivatives 810 The Chain Rule 816 Directional Derivatives and Gradient Vectors 824 Tangent Planes and Differentials 829 Extreme Values and Saddle Points 836 Lagrange Multipliers 849 Taylor's Formula for Two Variables 857 Partial Derivatives with Constrained Variables 859 Practice Exercises 862 Additional Exercises 876 15 Multiple Integrals 881 15.1 15.2 15.3 15.4 15.5 15.6 15.7 15.8 Double and Iterated Integrals over Rectangles 881 Double Integrals over General Regions 882 Area by Double Integration 896 Double Integrals in Polar Form 900 Triple Integrals in Rectangular Coordinates 904 Moments and Centers of Mass 909 Triple Integrals in Cylindrical and Spherical Coordinates 914 Substitutions in Multiple Integrals 922 Practice Exercises 927 Additional Exercises 933 16 Integration in Vector Fields 939 16.1 16.2 16.3 16.4 16.5 16.6 16.7 16.8 Line Integrals 939 Vector Fields and Line Integrals; Work, Circulation, and Flux 944 Path Independence, Potential Functions, and Conservative Fields 952 Green's Theorem in the Plane 957 Surfaces and Area 963 Surface Integrals 972 Stokes's Theorem 980 The Divergence Theorem and a Unified Theory 984 Practice Exercises 989 Additional Exercises 997 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. CHAPTER 1 FUNCTIONS 1.1 FUNCTIONS AND THEIR GRAPHS 1. domain œ (
_ß _); range œ [1ß _) 2. domain œ [0ß _); range œ (
_ß 1] 3. domain œ Ò
2ß _); y in range and y œ È5x  10 ! Ê y can be any positive real number Ê range œ Ò!ß _). 4. domain œ (
_ß 0Ó  Ò3, _); y in range and y œ Èx2
3x 5. domain œ (
_ß 3Ñ  Ð3, _); y in range and y œ Ê3
t!Ê 4 3
t 4 3
t, ! Ê y can be any positive real number Ê range œ Ò!ß _). now if t  3 Ê 3
t  ! Ê 4 3
t  !, or if t  3  ! Ê y can be any nonzero real number Ê range œ Ð
_ß 0Ñ  Ð!ß _). 6. domain œ (
_ß
%Ñ  Ð
4, 4Ñ  Ð4, _); y in range and y œ 2
%  t  4 Ê
16 Ÿ t
16  ! Ê nonzero real number Ê range œ Ð
_ß #
"'
18 Ó Ÿ 2 t2
16 2 t2
16 , 2 t2
16 now if t 
% Ê t2
16  ! Ê 2  !, or if t  % Ê t
16  ! Ê 2 t2
16  !, or if  ! Ê y can be any  Ð!ß _). 7. (a) Not the graph of a function of x since it fails the vertical line test. (b) Is the graph of a function of x since any vertical line intersects the graph at most once. 8. (a) Not the graph of a function of x since it fails the vertical line test. (b) Not the graph of a function of x since it fails the vertical line test. # 9. base œ x; (height)#  ˆ #x ‰ œ x# Ê height œ È3 # x; area is a(x) œ " # (base)(height) œ " # (x) Š È3 # x‹ œ È3 4 x# ; perimeter is p(x) œ x  x  x œ 3x. 10. s œ side length Ê s#  s# œ d# Ê s œ d È2 ; and area is a œ s# Ê a œ " # d# 11. Let D œ diagonal length of a face of the cube and j œ the length of an edge. Then j#  D# œ d# and D# œ 2j# Ê 3j# œ d# Ê j œ d È3 . The surface area is 6j# œ 6d# 3 12. The coordinates of P are ˆxß Èx‰ so the slope of the line joining P to the origin is m œ ˆx, Èx‰ œ ˆ m"# , # œ 2d# and the volume is j$ œ Š d3 ‹ Èx x œ " Èx $Î# œ (x  0). Thus, "‰ m . 13. 2x  4y œ 5 Ê y œ
"# x  54 ; L œ ÈÐx
0Ñ2  Ðy
0Ñ2 œ Éx2  Ð
"# x  54 Ñ2 œ Éx2  4" x2
54 x  œ É 54 x2
54 x  25 16 œ É 20x 2
20x  25 16 œ È20x2
20x  25 4 14. y œ Èx
3 Ê y2  3 œ x; L œ ÈÐx
4Ñ2  Ðy
0Ñ2 œ ÈÐy2  3
4Ñ2  y2 œ ÈÐy2
1Ñ2  y2 œ Èy4
2y2  1  y2 œ Èy4
y2  1 Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. d$ 3È 3 25 16 . 2 Chapter 1 Functions 15. The domain is a
_ß _b. 16. The domain is a
_ß _b. 17. The domain is a
_ß _b. 18. The domain is Ð
_ß !Ó. 19. The domain is a
_ß !b  a!ß _b. 20. The domain is a
_ß !b  a!ß _b. 21. The domain is a
_ß
5b  Ð
5ß
3Ó  Ò3, 5Ñ  a5, _b 22. The range is Ò2, 3Ñ. 23. Neither graph passes the vertical line test (a) (b) Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Section 1.1 Functions and Their Graphs 24. Neither graph passes the vertical line test (a) (b) Ú xyœ" Þ Ú yœ1
x Þ or or kx  yk œ 1 Í Û Í Û ß ß Ü x  y œ
" à Ü y œ
"
x à 25. x y 0 0 1 1 27. Faxb œ œ 2 0 26. x y 0 1 1 0 2 0 " , x0 28. Gaxb œ œ x x, 0 Ÿ x 4
x2 , x Ÿ 1 x2  2x, x  1 29. (a) Line through a!ß !b and a"ß "b: y œ x; Line through a"ß "b and a#ß !b: y œ
x  2 x, 0 Ÿ x Ÿ 1 f(x) œ œ
x  2, 1  x Ÿ 2 Ú Ý 2, ! Ÿ x  " Ý !ß " Ÿ x  # (b) f(x) œ Û Ý Ý 2ß # Ÿ x  $ Ü !ß $ Ÿ x Ÿ % 30. (a) Line through a!ß 2b and a#ß !b: y œ
x  2 " Line through a2ß "b and a&ß !b: m œ !&

# œ
x  #, 0  x Ÿ # f(x) œ œ "
$ x  &$ , #  x Ÿ & f(x) œ œ œ
"$ , so y œ
"$ ax
2b  " œ
"$ x 
$
! !
Ð
"Ñ œ
"
$
% #
! œ # (b) Line through a
"ß !b and a!ß
$b: m œ Line through a!ß $b and a#ß
"b: m œ
" $ & $
$, so y œ
$x
$ œ
#, so y œ
#x  $
$x
$,
"  x Ÿ !
#x  $, !  x Ÿ # Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 3 4 Chapter 1 Functions 31. (a) Line through a
"ß "b and a!ß !b: y œ
x Line through a!ß "b and a"ß "b: y œ " Line through a"ß "b and a$ß !b: m œ !
" $
" œ Ú
x
" Ÿ x  ! " !xŸ" f(x) œ Û Ü
"# x  $# "x$
" # (b) Line through a
2ß
1b and a0ß 0b: y œ 12 x Line through a0ß 2b and a1ß 0b: y œ
2x  2 Line through a1ß
1b and a3ß
1b: y œ
1 32. (a) Line through ˆ T# ß !‰ and aTß "b: m œ faxb œ
(b) "
! T
aTÎ#b œ
"# , so y œ
"# ax
"b  " œ
"# x  Ú 1 2x faxb œ Û
2x  2 Ü
1 $ #
2 Ÿ x Ÿ 0 0xŸ1 1xŸ3 œ T# , so y œ T# ˆx
T# ‰  0 œ T# x
" !, 0 Ÿ x Ÿ T# # T T x
", #  x Ÿ T Ú A, Ý Ý Ý
Aß faxb œ Û Aß Ý Ý Ý Ü
Aß ! Ÿ x  T# T # Ÿx T T Ÿ x  $#T $T # Ÿ x Ÿ #T 33. (a) ÚxÛ œ 0 for x − [0ß 1) (b) ÜxÝ œ 0 for x − (
1ß 0] 34. ÚxÛ œ ÜxÝ only when x is an integer. 35. For any real number x, n Ÿ x Ÿ n  ", where n is an integer. Now: n Ÿ x Ÿ n  " Ê
Ðn  "Ñ Ÿ
x Ÿ
n. By definition: Ü
xÝ œ
n and ÚxÛ œ n Ê
ÚxÛ œ
n. So Ü
xÝ œ
ÚxÛ for all x − d . 36. To find f(x) you delete the decimal or fractional portion of x, leaving only the integer part. Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Section 1.1 Functions and Their Graphs 37. Symmetric about the origin Dec:
_  x  _ Inc: nowhere 38. Symmetric about the y-axis Dec:
_  x  ! Inc: !  x  _ 39. Symmetric about the origin Dec: nowhere Inc:
_  x  ! !x_ 40. Symmetric about the y-axis Dec: !  x  _ Inc:
_  x  ! 41. Symmetric about the y-axis Dec:
_  x Ÿ ! Inc: !  x  _ 42. No symmetry Dec:
_  x Ÿ ! Inc: nowhere Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 5 6 Chapter 1 Functions 43. Symmetric about the origin Dec: nowhere Inc:
_  x  _ 44. No symmetry Dec: ! Ÿ x  _ Inc: nowhere 45. No symmetry Dec: ! Ÿ x  _ Inc: nowhere 46. Symmetric about the y-axis Dec:
_  x Ÿ ! Inc: !  x  _ 47. Since a horizontal line not through the origin is symmetric with respect to the y-axis, but not with respect to the origin, the function is even. 48. faxb œ x
& œ " x& and fa
xb œ a
xb
& œ " a
x b& œ
ˆ x"& ‰ œ
faxb. Thus the function is odd. 49. Since faxb œ x#  " œ a
xb#  " œ
faxb. The function is even. 50. Since Òfaxb œ x#  xÓ Á Òfa
xb œ a
xb#
xÓ and Òfaxb œ x#  xÓ Á Ò
faxb œ
axb#
xÓ the function is neither even nor odd. 51. Since gaxb œ x$  x, ga
xb œ
x$
x œ
ax$  xb œ
gaxb. So the function is odd. 52. gaxb œ x%  $x#
" œ a
xb%  $a
xb#
" œ ga
xbß thus the function is even. 53. gaxb œ " x#
" 54. gaxb œ x x#
" ; 55. hatb œ " t
"; œ " a
xb#
" œ ga
xb. Thus the function is even. ga
xb œ
x#x
" œ
gaxb. So the function is odd. ha
tb œ "
t
" ;
h at b œ " "
t. Since hatb Á
hatb and hatb Á ha
tb, the function is neither even nor odd. 56. Since l t$ | œ l a
tb$ |, hatb œ ha
tb and the function is even. Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Section 1.1 Functions and Their Graphs 57. hatb œ 2t  ", ha
tb œ
2t  ". So hatb Á ha
tb.
hatb œ
2t
", so hatb Á
hatb. The function is neither even nor odd. 58. hatb œ 2l t l  " and ha
tb œ 2l
t l  " œ 2l t l  ". So hatb œ ha
tb and the function is even. 59. s œ kt Ê 25 œ kÐ75Ñ Ê k œ " 3 Ê s œ 3" t; 60 œ 3" t Ê t œ 180 60. K œ c v# Ê 12960 œ ca18b2 Ê c œ 40 Ê K œ 40v# ; K œ 40a10b# œ 4000 joules 61. r œ 62. P œ k s Ê6œ k v k 4 Ê k œ 24 Ê r œ Ê 14.7 œ k 1000 24 s ; 10 œ 24 s Ê k œ 14700 Ê P œ Êsœ 14700 v ; 12 5 23.4 œ 14700 v Êvœ 24500 39 ¸ 628.2 in3 63. v œ f(x) œ xÐ"%
2xÑÐ22
2xÑ œ %x$
72x#  $!)x; !  x  7Þ 64. (a) Let h œ height of the triangle. Since the triangle is isosceles, AB #  AB # œ 2# Ê AB œ È2Þ So, # h#  "# œ ŠÈ2‹ Ê h œ " Ê B is at a!ß "b Ê slope of AB œ
" Ê The equation of AB is y œ f(x) œ
x  "; x − Ò!ß "Ó. (b) AÐxÑ œ 2x y œ 2xÐ
x  "Ñ œ
2x#  #x; x − Ò!ß "Ó. 65. (a) Graph h because it is an even function and rises less rapidly than does Graph g. (b) Graph f because it is an odd function. (c) Graph g because it is an even function and rises more rapidly than does Graph h. 66. (a) Graph f because it is linear. (b) Graph g because it contains a!ß "b. (c) Graph h because it is a nonlinear odd function. x # 67. (a) From the graph, (b) x # 1 x  0: x # x  0: x 2 4 x
1
Ê 4 x x #  1 4 x Ê x − (
2ß 0)  (%ß _)
1
4x  0 # 2x
8 0 Ê x
2x 0 Ê (x
4)(x2) #x 0 (x
4)(x2) #x 0 Ê x  4 since x is positive;
1
4 x 0 Ê x#
2x
8 2x 0 Ê Ê x 
2 since x is negative; sign of (x
4)(x  2)  
ïïïïïðïïïïïðïïïïî
2 % Solution interval: (
#ß 0)  (%ß _) Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 7 8 Chapter 1 Functions 3 2 x
1  x  1 3 2 x
1  x  1 68. (a) From the graph, (b) Case x 
1: Ê x − (
_ß
5)  (
1ß 1) Ê 3(x1) x
1 2 Ê 3x  3  2x
2 Ê x 
5. Thus, x − (
_ß
5) solves the inequality. Case
1  x  1: 3 x
1  2 x 1 3(x1) x
1 Ê 2 Ê 3x  3  2x
2 Ê x 
5 which is true if x 
1. Thus, x − (
1ß 1) solves the inequality. 3 Case 1  x: x
1  x2 1 Ê 3x  3  2x
2 Ê x 
5 which is never true if 1  x, so no solution here. In conclusion, x − (
_ß
5)  (
1ß 1). 69. A curve symmetric about the x-axis will not pass the vertical line test because the points ax, yb and ax,
yb lie on the same vertical line. The graph of the function y œ faxb œ ! is the x-axis, a horizontal line for which there is a single y-value, !, for any x. 70. price œ 40  5x, quantity œ 300
25x Ê Raxb œ a40  5xba300
25xb 71. x2  x2 œ h2 Ê x œ h È2 œ È2 h 2 ; cost œ 5a2xb  10h Ê Cahb œ 10Š È2 h 2 ‹  10h œ 5hŠÈ2  2‹ 72. (a) Note that 2 mi = 10,560 ft, so there are È800#  x# feet of river cable at $180 per foot and a10,560
xb feet of land cable at $100 per foot. The cost is Caxb œ 180È800#  x#  100a10,560
xb. (b) Ca!b œ $"ß #!!ß !!! Ca&!!b ¸ $"ß "(&ß )"# Ca"!!!b ¸ $"ß ")'ß &"# Ca"&!!b ¸ $"ß #"#ß !!! Ca#!!!b ¸ $"ß #%$ß ($# Ca#&!!b ¸ $"ß #()ß %(* Ca$!!!b ¸ $"ß $"%ß )(! Values beyond this are all larger. It would appear that the least expensive location is less than 2000 feet from the point P. 1.2 COMBINING FUNCTIONS; SHIFTING AND SCALING GRAPHS 1. Df :
_  x  _, Dg : x 2. Df : x  1 Rf œ Rg : y 0 Ê x 0, Rf g : y 1 Ê Df g
1, Dg : x
1 È2, Rfg : y œ Dfg : x 0 Ê x 1. Rf :
_  y  _, Rg : y 1. Therefore Df g œ Dfg : x 0, Rf g : y 1, Rfg : y 1. 0 3. Df :
_  x  _, Dg :
_  x  _, DfÎg :
_  x  _, DgÎf :
_  x  _, Rf : y œ 2, Rg : y RfÎg : 0  y Ÿ 2, RgÎf : "# Ÿ y  _ 4. Df :
_  x  _, Dg : x 0 , DfÎg : x 5. (a) 2 (d) (x  5)#
3 œ x#  10x  22 (g) x  10 0 0, DgÎf : x 0; Rf : y œ 1, Rg : y (b) 22 (e) 5 (h) (x#
3)#
3 œ x%
6x#  6 1, 1, RfÎg : 0  y Ÿ 1, RgÎf : 1 Ÿ y  _ (c) x#  2 (f)
2 Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Section 1.2 Combining Functions; Shifting and Scaling Graphs 6. (a)
"3 (d) (b) 2 " x (c) (e) 0 (g) x
2 (h) (f) " " x 1 1 œ x x " # 1 x" x# œ " x 1 3 4
1œ
x x1 7. af‰g‰hbaxb œ fagahaxbbb œ faga4
xbb œ fa3a4
xbb œ fa12
3xb œ a12
3xb  1 œ 13
3x 8. af‰g‰hbaxb œ fagahaxbbb œ fagax2 bb œ fa2ax2 b
1b œ fa2x2
1b œ 3a2x2
1b  4 œ 6x2 1 9. af‰g‰hbaxb œ fagahaxbbb œ fˆgˆ 1x ‰‰ œ fŠ 1 1 % ‹ œ fˆ 1 x 4x ‰ œ É 1 x 4x  " œ É 15x4x" x 2 10. af‰g‰hbaxb œ fagahaxbbb œ fŠgŠÈ2
x‹‹ œ f ŠÈ2
x‹ 2 ŠÈ2
x‹  œ fˆ $
x ‰ œ 1 2
x 2 x $ x 2 3
$2 xx 8
3x 7
2x œ 11. (a) af‰gbaxb (d) a j‰jbaxb (b) a j‰gbaxb (e) ag‰h‰f baxb (c) ag‰gbaxb (f) ah‰j‰f baxb 12. (a) af‰jbaxb (d) af‰f baxb (b) ag‰hbaxb (e) a j‰g‰f baxb (c) ah‰hbaxb (f) ag‰f‰hbaxb g(x) f(x) (f ‰ g)(x) (a) x
7 Èx Èx
7 (b) x2 3x 3(x  2) œ 3x  6 (c) x# Èx
5 Èx#
5 (d) x x
1 x x
1 " x
1 " x 1 13. (e) (f) " x gaxb
" g ax b œ x x
(x
1) œx x " x x " lx
"l . 14. (a) af‰gbaxb œ lgaxbl œ (b) af‰gbaxb œ x x 1 x x 1
1 x x" œ Ê"
" g ax b œ x x" Ê"
x x" œ " g ax b Ê " x" œ " gaxb ß so gaxb œ x  ". (c) Since af‰gbaxb œ Ègaxb œ lxl, gaxb œ x . (d) Since af‰gbaxb œ fˆÈx‰ œ l x l, faxb œ x# . (Note that the domain of the composite is Ò!ß _Ñ.) # The completed table is shown. Note that the absolute value sign in part (d) is optional. gaxb faxb af‰gbaxb " " lxl x
" lx
"l x" x# Èx x
" x Èx # x 15. (a) faga
1bb œ fa1b œ 1 (d) gaga2bb œ ga0b œ 0 x x" lxl lxl (b) gafa0bb œ ga
2b œ 2 (e) gafa
2bb œ ga1b œ
1 (c) fafa
1bb œ fa0b œ
2 (f) faga1bb œ fa
1b œ 0 16. (a) faga0bb œ fa
1b œ 2
a
1b œ 3, where ga0b œ 0
1 œ
1 (b) gafa3bb œ ga
1b œ
a
1b œ 1, where fa3b œ 2
3 œ
1 (c) gaga
1bb œ ga1b œ 1
1 œ 0, where ga
1b œ
a
1b œ 1 Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 9 10 Chapter 1 Functions (d) fafa2bb œ fa0b œ 2
0 œ 2, where fa2b œ 2
2 œ 0 (e) gafa0bb œ ga2b œ 2
1 œ 1, where fa0b œ 2
0 œ 2 (f) fˆgˆ "# ‰‰ œ fˆ
#" ‰ œ 2
ˆ
#" ‰ œ 5# , where gˆ "# ‰ œ "#
1 œ
"# 17. (a) af‰gbaxb œ fagaxbb œ É 1x  1 œ É 1 x x ag‰f baxb œ gafaxbb œ 1 Èx  1 (b) Domain af‰gb: Ð
_,
1Ó  Ð0, _Ñ, domain ag‰f b: Ð
1, _Ñ (c) Range af‰gb: Ð1, _Ñ, range ag‰f b: Ð0, _Ñ 18. (a) af‰gbaxb œ fagaxbb œ 1
2Èx  x ag‰f baxb œ gafaxbb œ 1
kxk (b) Domain af‰gb: Ò0, _Ñ, domain ag‰f b: Ð
_, _Ñ (c) Range af‰gb: Ð0, _Ñ, range ag‰f b: Ð
_, 1Ó 19. af‰gbaxb œ x Ê fagaxbb œ x Ê g ax b g ax b
2 œ x Ê gaxb œ agaxb
2bx œ x † gaxb
2x Ê gaxb
x † gaxb œ
2x Ê gaxb œ
1 2x
x œ 2x x
1 20. af‰gbaxb œ x  2 Ê fagaxbb œ x  2 Ê 2agaxbb3
4 œ x  2 Ê agaxbb3 œ 21. (a) y œ
(x  7)# (b) y œ
(x
4)# 22. (a) y œ x#  3 (b) y œ x#
5 x6 2 3 x6 Ê gaxb œ É 2 23. (a) Position 4 (b) Position 1 (c) Position 2 (d) Position 3 24. (a) y œ
(x
1)#  4 (b) y œ
(x  2)#  3 (c) y œ
(x  4)#
1 (d) y œ
(x
2)# 25. 26. 27. 28. Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Section 1.2 Combining Functions; Shifting and Scaling Graphs 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 11 12 Chapter 1 Functions 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Section 1.2 Combining Functions; Shifting and Scaling Graphs 53. 54. 55. (a) domain: [0ß 2]; range: [#ß $] (b) domain: [0ß 2]; range: [
1ß 0] (c) domain: [0ß 2]; range: [0ß 2] (d) domain: [0ß 2]; range: [
1ß 0] (e) domain: [
2ß 0]; range: [!ß 1] (f) domain: [1ß 3]; range: [!ß "] (g) domain: [
2ß 0]; range: [!ß "] (h) domain: [
1ß 1]; range: [!ß "] Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 13 14 Chapter 1 Functions 56. (a) domain: [0ß 4]; range: [
3ß 0] (b) domain: [
4ß 0]; range: [!ß $] (c) domain: [
4ß 0]; range: [!ß $] (d) domain: [
4ß 0]; range: ["ß %] (e) domain: [#ß 4]; range: [
3ß 0] (f) domain: [
2ß 2]; range: [
3ß 0] (g) domain: ["ß 5]; range: [
3ß 0] (h) domain: [0ß 4]; range: [0ß 3] 58. y œ a2xb#
1 œ %x#
1 57. y œ 3x#
3 59. y œ "# ˆ"  "‰ x# œ " #  " #x# 60. y œ 1  " axÎ$b# œ1 61. y œ È%x  1 62. y œ 3Èx  1 # 63. y œ É%
ˆ x# ‰ œ "# È16
x# 64. y œ "$ È%
x# 65. y œ "
a3xb$ œ "
27x$ 66. y œ "
ˆ x# ‰ œ "
$ * x# x$ ) Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Section 1.2 Combining Functions; Shifting and Scaling Graphs 67. Let y œ
È#x  " œ faxb and let gaxb œ x"Î# , "Î# "Î# haxb œ ˆx  " ‰ , iaxb œ È#ˆx  " ‰ , and # # "Î# jaxb œ
’È#ˆx  "# ‰ “ œ faBb. The graph of haxb is the graph of gaxb shifted left " # unit; the graph of iaxb is the graph of haxb stretched vertically by a factor of È#; and the graph of jaxb œ faxb is the graph of iaxb reflected across the x-axis. 68. Let y œ È"
x # œ faxbÞ Let gaxb œ a
xb"Î# , haxb œ a
x  #b"Î# , and iaxb œ œ È"
x # " È # a
x  #b"Î# œ faxbÞ The graph of gaxb is the graph of y œ Èx reflected across the x-axis. The graph of haxb is the graph of gaxb shifted right two units. And the graph of iaxb is the graph of haxb compressed vertically by a factor of È#. 69. y œ faxb œ x$ . Shift faxb one unit right followed by a shift two units up to get gaxb œ ax
"b3  #. 70. y œ a"
Bb$  # œ
Òax
"b$  a
#bÓ œ faxb. Let gaxb œ x$ , haxb œ ax
"b$ , iaxb œ ax
"b$  a
#b, and jaxb œ
Òax
"b$  a
#bÓ. The graph of haxb is the graph of gaxb shifted right one unit; the graph of iaxb is the graph of haxb shifted down two units; and the graph of faxb is the graph of iaxb reflected across the x-axis. 71. Compress the graph of faxb œ of 2 to get gaxb œ unit to get haxb œ " #x . Then " #x
". " x horizontally by a factor shift gaxb vertically down 1 Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 15 16 Chapter 1 Functions 72. Let faxb œ œ " # ŠxÎÈ#‹ " x# and gaxb œ "œ # x# "œ " # ’Š"ÎÈ#‹B“ " # Š B# ‹ "  "Þ Since È# ¸ "Þ%, we see that the graph of faxb stretched horizontally by a factor of 1.4 and shifted up 1 unit is the graph of gaxb. $ 73. Reflect the graph of y œ faxb œ È x across the x-axis $ to get gaxb œ
Èx. 74. y œ faxb œ a
#xb#Î$ œ Òa
"ba#bxÓ#Î$ œ a
"b#Î$ a#xb#Î$ œ a#xb#Î$ . So the graph of faxb is the graph of gaxb œ x#Î$ compressed horizontally by a factor of 2. 75. 76. 77. *x#  #&y# œ ##& Ê x# &#  y# $# œ" 78. "'x#  (y# œ ""# Ê x# # È Š (‹  y# %# œ" Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Section 1.2 Combining Functions; Shifting and Scaling Graphs 79. $x#  ay
#b# œ $ Ê x# "#  a y
#b # # ŠÈ$‹ 80. ax  "b#  #y# œ % Ê œ" Ê 83. x# "' # ŠÈ#‹  y# * 
y
a
#b‘# # ŠÈ$‹  # # 82. 'ˆx  $# ‰  *ˆy
"# ‰ œ &% 81. $ax
"b#  #ay  #b# œ ' ax
" b #
x
a
"b‘# ## # œ" Ê ’x
ˆ
$# ‰“ $#  ˆy
"# ‰# # ŠÈ'‹ œ" œ " has its center at a!ß !b. Shiftinig 4 units left and 3 units up gives the center at ah, kb œ a
%ß $b. #
x
a
4b‘#  ay
3#3b œ " 4# a y
$b # œ ". Center, C, is a
%ß 3# So the equation is Ê ax  % b # 4#  $b, and major axis, AB, is the segment from a
)ß $b to a!ß $b. 84. The ellipse x# % y# #&  œ " has center ah, kb œ a!ß !b. Shifting the ellipse 3 units right and 2 units down produces an ellipse with center at ah, kb œ a$ß
#b and an equation ax
3 b# % 
y
a
#b‘# #& œ ". Center, C, is a3ß
#b, and AB, the segment from a$ß $b to a$ß
(b is the major axis. 85. (a) (fg)(
x) œ f(
x)g(
x) œ f(x)(
g(x)) œ
(fg)(x), odd (b) Š gf ‹ (
x) œ f(
x) g(
x) œ f(x)
g(x) œ
Š gf ‹ (x), odd Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. y# # È Š #‹ œ" 17 18 Chapter 1 Functions (c) ˆ gf ‰ (
x) œ (d) (e) (f) (g) (h) (i) g(
x) f(
x) œ
g(x) f(x) œ
ˆ gf ‰ (x), odd f # (
x) œ f(
x)f(
x) œ f(x)f(x) œ f # (x), even g# (
x) œ (g(
x))# œ (
g(x))# œ g# (x), even (f ‰ g)(
x) œ f(g(
x)) œ f(
g(x)) œ f(g(x)) œ (f ‰ g)(x), even (g ‰ f)(
x) œ g(f(
x)) œ g(f(x)) œ (g ‰ f)(x), even (f ‰ f)(
x) œ f(f(
x)) œ f(f(x)) œ (f ‰ f)(x), even (g ‰ g)(
x) œ g(g(
x)) œ g(
g(x)) œ
g(g(x)) œ
(g ‰ g)(x), odd 86. Yes, f(x) œ 0 is both even and odd since f(
x) œ 0 œ f(x) and f(
x) œ 0 œ
f(x). 87. (a) (b) (c) (d) 88. Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Section 1.3 Trigonometric Functions 1.3 TRIGONOMETRIC FUNCTIONS 1. (a) s œ r) œ (10) ˆ 451 ‰ œ 81 m radians and 51 4 1 ‰ 3. ) œ 80° Ê ) œ 80° ˆ 180° œ 41 9 2. ) œ s r œ 101 8 œ 51 4 1 ‰ (b) s œ r) œ (10)(110°) ˆ 180° œ
1 )
231 0 1 # s r œ 30 50 31 4 " È2
È" 2 sin ) 0 cos )
1 tan ) 0 È3 0 und.
" und. " È3 und. 0
1 und.
È 2
1
# und.
È23 sec ) csc ) 0 " " 0 " und. 7. cos x œ
45 , tan x œ
34 9. sin x œ
È8 3 , tan x œ
È8 " 6. È2
3#1 )
1' sin ) " cos ) ! " # tan ) und.
È 3 cot ) !
È"3 sec ) und. # csc ) "
È23 8. sin x œ 2 È5 10. sin x œ 12 13 13. 14. period œ 1
13 È
#3 12. cos x œ
, cos x œ " È2 &1 ' " # È
#3
È"3 "
È"3
È 3 "
È 3 2 È3 È2
È23
# È2 #
"# È3 # " È5 , tan x œ
12 5 È3 # , tan x œ " È3 period œ 41 16. period œ 2 m ‰ ¸ 34° œ 0.6 rad or 0.6 ˆ 180° 1 11. sin x œ
È"5 , cos x œ
È25 15. 551 9 Ê s œ (6) ˆ 491 ‰ œ 8.4 in. (since the diameter œ 12 in. Ê radius œ 6 in.) È
#3
"# cot ) œ ˆ 180° ‰ œ 225° 1 4. d œ 1 meter Ê r œ 50 cm Ê ) œ 5. 1101 18 period œ 4 Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 1 % " È2 19 20 Chapter 1 Functions 17. 18. period œ 6 period œ 1 19. 20. period œ 21 period œ 21 21. 22. period œ 21 period œ 21 23. period œ 1# , symmetric about the origin 24. period œ 1, symmetric about the origin 25. period œ 4, symmetric about the s-axis 26. period œ 41, symmetric about the origin Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Section 1.3 Trigonometric Functions 27. (a) Cos x and sec x are positive for x in the interval ˆ
12 , 12 ‰; and cos x and sec x are negative for x in the intervals ˆ
321 ,
12 ‰ and ˆ 12 , 321 ‰. Sec x is undefined when cos x is 0. The range of sec x is (
_ß
1]  ["ß _); the range of cos x is [
"ß 1]. (b) Sin x and csc x are positive for x in the intervals ˆ
321 ,
1‰ and a!, 1b; and sin x and csc x are negative for x in the intervals a
1, !b and ˆ1, 321 ‰. Csc x is undefined when sin x is 0. The range of csc x is (
_ß
1]  [1ß _); the range of sin x is [
"ß "]. 28. Since cot x œ " tan x , cot x is undefined when tan x œ 0 and is zero when tan x is undefined. As tan x approaches zero through positive values, cot x approaches infinity. Also, cot x approaches negative infinity as tan x approaches zero through negative values. 29. D:
_  x  _; R: y œ
1, 0, 1 30. D:
_  x  _; R: y œ
1, 0, 1 31. cos ˆx
1# ‰ œ cos x cos ˆ
1# ‰
sin x sin ˆ
1# ‰ œ (cos x)(0)
(sin x)(
1) œ sin x 32. cos ˆx  1# ‰ œ cos x cos ˆ 1# ‰
sin x sin ˆ 1# ‰ œ (cos x)(0)
(sin x)(1) œ
sin x 33. sin ˆx  1# ‰ œ sin x cos ˆ 1# ‰  cos x sin ˆ 1# ‰ œ (sin x)(0)  (cos x)(1) œ cos x 34. sin ˆx
1# ‰ œ sin x cos ˆ
1# ‰  cos x sin ˆ
1# ‰ œ (sin x)(0)  (cos x)(
1) œ
cos x 35. cos (A
B) œ cos (A  (
B)) œ cos A cos (
B)
sin A sin (
B) œ cos A cos B
sin A (
sin B) œ cos A cos B  sin A sin B 36. sin (A
B) œ sin (A  (
B)) œ sin A cos (
B)  cos A sin (
B) œ sin A cos B  cos A (
sin B) œ sin A cos B
cos A sin B 37. If B œ A, A
B œ 0 Ê cos (A
B) œ cos 0 œ 1. Also cos (A
B) œ cos (A
A) œ cos A cos A  sin A sin A œ cos# A  sin# A. Therefore, cos# A  sin# A œ 1. Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 21 22 Chapter 1 Functions 38. If B œ 21, then cos (A  21) œ cos A cos 21
sin A sin 21 œ (cos A)(1)
(sin A)(0) œ cos A and sin (A  21) œ sin A cos 21  cos A sin 21 œ (sin A)(1)  (cos A)(0) œ sin A. The result agrees with the fact that the cosine and sine functions have period 21. 39. cos (1  x) œ cos 1 cos B
sin 1 sin x œ (
1)(cos x)
(0)(sin x) œ
cos x 40. sin (21
x) œ sin 21 cos (
x)  cos (21) sin (
x) œ (0)(cos (
x))  (1)(sin (
x)) œ
sin x 41. sin ˆ 3#1
x‰ œ sin ˆ 3#1 ‰ cos (
x)  cos ˆ 3#1 ‰ sin (
x) œ (
1)(cos x)  (0)(sin (
x)) œ
cos x 42. cos ˆ 3#1  x‰ œ cos ˆ 3#1 ‰ cos x
sin ˆ 3#1 ‰ sin x œ (0)(cos x)
(
1)(sin x) œ sin x œ sin ˆ 14  13 ‰ œ sin 44. cos 111 1# 45. cos 1 12 œ cos ˆ 13
14 ‰ œ cos 46. sin 51 1# œ sin ˆ 231
14 ‰ œ sin ˆ 231 ‰ cos ˆ
14 ‰  cos ˆ 231 ‰ sin ˆ
14 ‰ œ Š 21 ‰ 3 È2 1 8 œ 1
cos ˆ 281 ‰ # œ 1
# 49. sin# 1 1# œ 1  cos ˆ 211# ‰ # œ 1 # 3 4 Ê sin ) œ „ 52. sin2 ) œ cos2 ) Ê sin2 ) cos2 ) 1 4 œ cos 47. cos# 51. sin2 ) œ cos # È3 # È3 2 œ 1 3  cos cos 21 3 1 4 sin
sin 1 4 cos ˆ
14 ‰
sin 1 3 1 3 È2 È3 # ‹Š # ‹ 71 1# œ cos ˆ 14  1 4 È2 ˆ"‰ # ‹ # 43. sin œŠ sin 1 3 21 3 œŠ Š È2 ˆ "‰ # ‹
# sin ˆ
14 ‰ œ ˆ "# ‰ Š
Š È2 # ‹ œ È2 È3 # ‹Š # ‹
Š È3 È2 # ‹Š # ‹ 2
È2 4 48. cos# 51 1# œ 1‰ 1
cos ˆ 10 1# # œ 2  È3 4 50. sin# 31 8 œ 1  cos ˆ 681 ‰ # Ê tan2 ) œ 1 Ê tan ) œ „ 1 Ê ) œ 14 , 31 51 71 4 , 4 , 4 cos2 ) cos2 ) œ
È3 È2 # ‹ Š
# ‹ œ Ê ) œ 13 , È 6
È 2 4 È 2
È 6 4 1
È 3 2È 2 œ  ˆ
"# ‰ Š
œ œ 1
Š È3 ‹ # # 1  Š # È2 ‹ # È2 # ‹ œ œ œ 1
È 3 2È 2 2  È3 4 2
È2 4 21 41 51 3 , 3 , 3 53. sin 2)
cos ) œ 0 Ê 2sin ) cos )
cos ) œ 0 Ê cos )a2sin )
1b œ 0 Ê cos ) œ 0 or 2sin )
1 œ 0 Ê cos ) œ 0 or sin ) œ "# Ê ) œ 12 , 321 , or ) œ 16 , 561 Ê ) œ 16 , 12 , 561 , 321 54. cos 2)  cos ) œ 0 Ê 2cos2 )
1  cos ) œ 0 Ê 2cos2 )  cos )
1 œ 0 Ê acos )  1ba2cos )
1b œ 0 Ê cos )  1 œ 0 or 2cos )
1 œ 0 Ê cos ) œ
1 or cos ) œ "# Ê ) œ 1 or ) œ 13 , 531 Ê ) œ 13 , 1, 531 55. tan (A  B) œ sin (A
B) cos (A
B) œ sin A cos B
cos A cos B cos A cos Bsin A sin B œ sin A cos B cos A sin B cos A cos B
cos A cos B sin A sin B cos A cos B  cos A cos B cos A cos B œ tan A
tan B 1tan A tan B 56. tan (A
B) œ sin (AB) cos (AB) œ sin A cos Bcos A cos B cos A cos B
sin A sin B œ sin A cos B cos A sin B cos A cos B  cos A cos B sin A sin B cos A cos B
cos A cos B cos A cos B œ tan Atan B 1
tan A tan B 57. According to the figure in the text, we have the following: By the law of cosines, c# œ a#  b#
2ab cos ) œ 1#  1#
2 cos (A
B) œ 2
2 cos (A
B). By distance formula, c# œ (cos A
cos B)#  (sin A
sin B)# œ cos# A
2 cos A cos B  cos# B  sin# A
2 sin A sin B  sin# B œ 2
2(cos A cos B  sin A sin B). Thus c# œ 2
2 cos (A
B) œ 2
2(cos A cos B  sin A sin B) Ê cos (A
B) œ cos A cos B  sin A sin B. Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Section 1.3 Trigonometric Functions 58. (a) cosaA
Bb œ cos A cos B  sin A sin B sin ) œ cosˆ 1#
)‰ and cos ) œ sinˆ 1#
)‰ Let ) œ A  B sinaA  Bb œ cos’ 1#
aA  Bb“ œ cos’ˆ 1#
A‰
B“ œ cos ˆ 1#
A‰ cos B  sin ˆ 1#
A‰ sin B œ sin A cos B  cos A sin B (b) cosaA
Bb œ cos A cos B  sin A sin B cosaA
a
Bbb œ cos A cos a
Bb  sin A sin a
Bb Ê cosaA  Bb œ cos A cos a
Bb  sin A sin a
Bb œ cos A cos B  sin A a
sin Bb œ cos A cos B
sin A sin B Because the cosine function is even and the sine functions is odd. 59. c# œ a#  b#
2ab cos C œ 2#  3#
2(2)(3) cos (60°) œ 4  9
12 cos (60°) œ 13
12 ˆ "# ‰ œ 7. Thus, c œ È7 ¸ 2.65. 60. c# œ a#  b#
2ab cos C œ 2#  3#
2(2)(3) cos (40°) œ 13
12 cos (40°). Thus, c œ È13
12 cos 40° ¸ 1.951. 61. From the figures in the text, we see that sin B œ hc . If C is an acute angle, then sin C œ hb . On the other hand, if C is obtuse (as in the figure on the right), then sin C œ sin (1
C) œ hb . Thus, in either case, h œ b sin C œ c sin B Ê ah œ ab sin C œ ac sin B. a #
b # c # 2ab By the law of cosines, cos C œ and cos B œ a #
c #  b # . 2ac Moreover, since the sum of the interior angles of a triangle is 1, we have sin A œ sin (1
(B  C)) œ sin (B  C) œ sin B cos C  cos B sin C # # # # # # b c c b ˆ h ‰ h ‰ œ ˆ hc ‰ ’ a
2ab a2a#  b#
c#  c#
b# b œ “  ’ a
2ac “ b œ ˆ 2abc ah bc Ê ah œ bc sin A. Combining our results we have ah œ ab sin C, ah œ ac sin B, and ah œ bc sin A. Dividing by abc gives h sin A sin C sin B bc œ ðóóóóóóóñóóóóóóóò a œ c œ b . law of sines 62. By the law of sines, sin A # œ sin B 3 œ È3/2 c . By Exercise 61 we know that c œ È7. Thus sin B œ 3È 3 2È 7 ¶ 0.982. 63. From the figure at the right and the law of cosines, b# œ a#  2#
2(2a) cos B œ a#  4
4a ˆ "# ‰ œ a#
2a  4. Applying the law of sines to the figure, Ê È2/2 a œ È3/2 b sin A a œ sin B b Ê b œ É 3# a. Thus, combining results, a#
2a  4 œ b# œ 3 # a# Ê 0 œ " # a#  2a
4 Ê 0 œ a#  4a
8. From the quadratic formula and the fact that a  0, we have aœ 4
È4# 4(1)(8) # œ 4 È 3 4 # ¶ 1.464. 64. (a) The graphs of y œ sin x and y œ x nearly coincide when x is near the origin (when the calculator is in radians mode). (b) In degree mode, when x is near zero degrees the sine of x is much closer to zero than x itself. The curves look like intersecting straight lines near the origin when the calculator is in degree mode. Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 23 24 Chapter 1 Functions 65. A œ 2, B œ 21, C œ
1, D œ
1 66. A œ "# , B œ 2, C œ 1, D œ " # 67. A œ
12 , B œ 4, C œ 0, D œ 68. A œ L 21 , " 1 B œ L, C œ 0, D œ 0 69-72. Example CAS commands: Maple f := x -> A*sin((2*Pi/B)*(x-C))+D1; A:=3; C:=0; D1:=0; f_list := [seq( f(x), B=[1,3,2*Pi,5*Pi] )]; plot( f_list, x=-4*Pi..4*Pi, scaling=constrained, color=[red,blue,green,cyan], linestyle=[1,3,4,7], legend=["B=1","B=3","B=2*Pi","B=3*Pi"], title="#69 (Section 1.3)" ); Mathematica Clear[a, b, c, d, f, x] f[x_]:=a Sin[21/b (x
c)] + d Plot[f[x]/.{a Ä 3, b Ä 1, c Ä 0, d Ä 0}, {x,
41, 41 }] Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Section 1.3 Trigonometric Functions 69. (a) The graph stretches horizontally. (b) The period remains the same: period œ l B l. The graph has a horizontal shift of " # period. 70. (a) The graph is shifted right C units. (b) The graph is shifted left C units. (c) A shift of „ one period will produce no apparent shift. l C l œ ' 71. (a) The graph shifts upwards l D lunits for D  ! (b) The graph shifts down l D lunits for D  !Þ 72. (a) The graph stretches l A l units. (b) For A  !, the graph is inverted. Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 25 26 Chapter 1 Functions 1.4 GRAPHING WITH CALCULATORS AND COMPUTERS 1-4. The most appropriate viewing window displays the maxima, minima, intercepts, and end behavior of the graphs and has little unused space. 1. d. 2. c. 3. d. 4. b. 5-30. For any display there are many appropriate display widows. The graphs given as answers in Exercises 5
30 are not unique in appearance. 5. Ò
2ß 5Ó by Ò
15ß 40Ó 6. Ò
4ß 4Ó by Ò
4ß 4Ó 7. Ò
2ß 6Ó by Ò
250ß 50Ó 8. Ò
1ß 5Ó by Ò
5ß 30Ó Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Section 1.4 Graphing with Calculators and Computers 9. Ò
4ß 4Ó by Ò
5ß 5Ó 10. Ò
2ß 2Ó by Ò
2ß 8Ó 11. Ò
2ß 6Ó by Ò
5ß 4Ó 12. Ò
4ß 4Ó by Ò
8ß 8Ó 13. Ò
1ß 6Ó by Ò
1ß 4Ó 14. Ò
1ß 6Ó by Ò
1ß 5Ó 15. Ò
3ß 3Ó by Ò0ß 10Ó 16. Ò
1ß 2Ó by Ò0ß 1Ó Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 27 28 Chapter 1 Functions 17. Ò
5ß 1Ó by Ò
5ß 5Ó 18. Ò
5ß 1Ó by Ò
2ß 4Ó 19. Ò
4ß 4Ó by Ò0ß 3Ó 20. Ò
5ß 5Ó by Ò
2ß 2Ó 21. Ò
"!ß "!Ó by Ò
'ß 'Ó 22. Ò
&ß &Ó by Ò
#ß #Ó 23. Ò
'ß "!Ó by Ò
'ß 'Ó 24. Ò
$ß &Ó by Ò
#ß "!Ó Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Section 1.4 Graphing with Calculators and Computers 25. Ò
0Þ03ß 0Þ03Ó by Ò
1Þ25ß 1Þ25Ó 26. Ò
0Þ1ß 0Þ1Ó by Ò
3ß 3Ó 27. Ò
300ß 300Ó by Ò
1Þ25ß 1Þ25Ó 28. Ò
50ß 50Ó by Ò
0Þ1ß 0Þ1Ó 29. Ò
0Þ25ß 0Þ25Ó by Ò
0Þ3ß 0Þ3Ó 30. Ò
0Þ15ß 0Þ15Ó by Ò
0Þ02ß 0Þ05Ó 31. x#  #x œ %  %y
y# Ê y œ # „ È
x#
#x  ). The lower half is produced by graphing y œ #
È
x#
#x  ). 32. y#
"'x# œ " Ê y œ „ È"  "'x# . The upper branch is produced by graphing y œ È"  "'x# . Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 29 30 Chapter 1 Functions 33. 34. 35. 36. 37. 38Þ 39. 40. CHAPTER 1 PRACTICE EXERCISES 1. The area is A œ 1 r# and the circumference is C œ #1 r. Thus, r œ 2. The surface area is S œ %1 r# Ê r œ ˆ %S1 ‰ surface area gives S œ %1 r# œ %1 ˆ $%V1 ‰ "Î# #Î$ C #1 # Ê A œ 1ˆ #C1 ‰ œ C# %1 . $ $V . The volume is V œ %$ 1 r$ Ê r œ É %1 . Substitution into the formula for . Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Chapter 1 Practice Exercises 3. The coordinates of a point on the parabola are axß x# b. The angle of inclination ) joining this point to the origin satisfies the equation tan ) œ 4. tan ) œ rise run œ h &!! x# x œ x. Thus the point has coordinates axß x# b œ atan )ß tan# )b. Ê h œ &!! tan ) ft. 5. 6. Symmetric about the origin. Symmetric about the y-axis. 7. 8. Neither Symmetric about the y-axis. 9. ya
xb œ a
xb#  " œ x#  " œ yaxb. Even. 10. ya
xb œ a
xb&
a
xb$
a
xb œ
x&  x$  x œ
yaxb. Odd. 11. ya
xb œ "
cosa
xb œ "
cos x œ yaxb. Even. 12. ya
xb œ seca
xb tana
xb œ 13. ya
xb œ a
xb% " a
xb$
#a
xb œ x% "
x$ #x sina
xb cos# a
xb œ
sin x cos# x œ
sec x tan x œ
yaxb. Odd. % " œ
xx$
# x œ
yaxb. Odd. 14. ya
xb œ a
xb
sina
xb œ a
xb  sin x œ
ax
sin xb œ
yaxb. Odd. 15. ya
xb œ
x  cosa
xb œ
x  cos x. Neither even nor odd. 16. ya
xb œ a
xbcosa
xb œ
x cos x œ
yaxb. Odd. 17. Since f and g are odd Ê fa
xb œ
faxb and ga
xb œ
gaxb. (a) af † gba
xb œ fa
xbga
xb œ Ò
faxbÓÒ
gaxbÓ œ faxbgaxb œ af † gbaxb Ê f † g is even (b) f 3 a
xb œ fa
xbfa
xbfa
xb œ Ò
faxbÓÒ
faxbÓÒ
faxbÓ œ
faxb † faxb † faxb œ
f 3 axb Ê f 3 is odd. (c) fasina
xbb œ fa
sinaxbb œ
fasinaxbb Ê fasinaxbb is odd. (d) gaseca
xbb œ gasecaxbb Ê gasecaxbb is even. (e) lga
xbl œ l
gaxbl œ lgaxbl Ê lgl is evenÞ Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 31 32 Chapter 1 Functions 18. Let faa
xb œ faa  xb and define gaxb œ fax  ab. Then ga
xb œ faa
xb  ab œ faa
xb œ faa  xb œ fax  ab œ gaxb Ê gaxb œ fax  ab is even. 19. (a) The function is defined for all values of x, so the domain is a
_ß _b. (b) Since l x l attains all nonnegative values, the range is Ò
#ß _Ñ. 20. (a) Since the square root requires "
x !, the domain is Ð
_ß "Ó. (b) Since È"
x attains all nonnegative values, the range is Ò
#ß _Ñ. 21. (a) Since the square root requires "'
x# !, the domain is Ò
%ß %Ó. (b) For values of x in the domain, ! Ÿ "'
x# Ÿ "', so ! Ÿ È"'
x# Ÿ %. The range is Ò!ß %Ó. 22. (a) The function is defined for all values of x, so the domain is a
_ß _b. (b) Since $#
x attains all positive values, the range is a"ß _b. 23. (a) The function is defined for all values of x, so the domain is a
_ß _b. (b) Since #e
x attains all positive values, the range is a
$ß _b. 24. (a) The function is equivalent to y œ tan #x, so we require #x Á k1 # for odd integers k. The domain is given by x Á k1 % for odd integers k. (b) Since the tangent function attains all values, the range is a
_ß _b. 25. (a) The function is defined for all values of x, so the domain is a
_ß _b. (b) The sine function attains values from
" to ", so
# Ÿ #sina$x  1b Ÿ # and hence
$ Ÿ #sina$x  1b
" Ÿ ". The range is Ò
3ß 1Ó. 26. (a) The function is defined for all values of x, so the domain is a
_ß _b. & (b) The function is equivalent to y œ È x# , which attains all nonnegative values. The range is Ò!ß _Ñ. 27. (a) The logarithm requires x
$  !, so the domain is a$ß _b. (b) The logarithm attains all real values, so the range is a
_ß _b. 28. (a) The function is defined for all values of x, so the domain is a
_ß _b. (b) The cube root attains all real values, so the range is a
_ß _b. 29. (a) (b) (c) (d) Increasing because volume increases as radius increases Neither, since the greatest integer function is composed of horizontal (constant) line segments Decreasing because as the height increases, the atmospheric pressure decreases. Increasing because the kinetic (motion) energy increases as the particles velocity increases. 30. (a) Increasing on Ò2, _Ñ (c) Increasing on a
_, _b (b) Increasing on Ò
1, _Ñ (d) Increasing on Ò "# , _Ñ 31. (a) The function is defined for
% Ÿ x Ÿ %, so the domain is Ò
%ß %Ó. (b) The function is equivalent to y œ Èl x l,
% Ÿ x Ÿ %, which attains values from ! to # for x in the domain. The range is Ò!ß #Ó. Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Chapter 1 Practice Exercises 33 32. (a) The function is defined for
# Ÿ x Ÿ #, so the domain is Ò
#ß #Ó. (b) The range is Ò
"ß "Ó. 33. First piece: Line through a!ß "b and a"ß !b. m œ Second piece: Line through a"ß "b and a#ß !b. m faxb œ œ "
x, ! Ÿ x  " #
x, " Ÿ x Ÿ # 34. First piece: Line through a!ß !b and a2ß 5b. m œ Second piece: Line through a2ß 5b and a4ß !b. m faxb œ
10
5 2 x, 5x 2 , !
"
" "
! œ " œ "
" œ !#

" œ "
" Ê y œ
x  " œ "
x œ
" Ê y œ
ax
"b  " œ
x  # œ #
x 5
! 5 5 2
! œ 2 Ê y œ 2x !
5
5 œ 4
2 œ 2 œ
52 Ê y œ
52 ax
2b  5 œ
52 x  10 œ 10
!Ÿx2 (Note: x œ 2 can be included on either piece.) 2ŸxŸ4 35. (a) af‰gba
"b œ faga
"bb œ fŠ È
""  # ‹ œ fa"b œ (b) ag‰f ba#b œ gafa#bb œ gˆ "2 ‰ œ (c) af‰f baxb œ fafaxbb œ fˆ "x ‰ œ " É "#  # " "Îx œ " È#Þ& " " œ" or É &# œ x, x Á ! (d) ag‰gbaxb œ gagaxbb œ gŠ Èx" # ‹ œ " " É Èx # # œ % x# È É "  #È x  # $ 36. (a) af‰gba
"b œ faga
"bb œ fˆÈ
"  "‰ œ fa!b œ #
! œ # $ (b) ag‰f ba#b œ faga#bb œ ga#
#b œ ga!b œ È !"œ" (c) af‰f baxb œ fafaxbb œ fa#
xb œ #
a#
xb œ x $ $ $ È (d) ag‰gbaxb œ gagaxbb œ gˆÈ x  "‰ œ É x"" # 37. (a) af‰gbaxb œ fagaxbb œ fˆÈx  #‰ œ #
ˆÈx  #‰ œ
x, x
#. ag‰f baxb œ fagaxbb œ ga#
x# b œ Èa#
x# b  # œ È%
x# (b) Domain of f‰g: Ò
#ß _ÑÞ Domain of g‰f: Ò
#ß #ÓÞ (c) Range of f‰g: Ð
_ß #ÓÞ Range of g‰f: Ò!ß #ÓÞ % 38. (a) af‰gbaxb œ fagaxbb œ fŠÈ"
x‹ œ ÉÈ"
x œ È "
x. ag‰f baxb œ fagaxbb œ gˆÈx‰ œ É"
Èx (b) Domain of f‰g: Ð
_ß "ÓÞ Domain of g‰f: Ò!ß "ÓÞ 39. y œ faxb (c) Range of f‰g: Ò!ß _ÑÞ Range of g‰f: Ò!ß "ÓÞ y œ af‰f baxb Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 5x 2 34 Chapter 1 Functions 40. 41. 42. The graph of f# (x) œ f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y œ f" (x), x 0 across the y-axis. 43. It does not change the graph. 44. Whenever g" (x) is positive, the graph of y œ g# (x) œ kg" (x)k is the same as the graph of y œ g" (x). When g" (x) is negative, the graph of y œ g# (x) is the reflection of the graph of y œ g" (x) across the x-axis. Whenever g" (x) is positive, the graph of y œ g# (x) œ kg" (x)k is the same as the graph of y œ g" (x). When g" (x) is negative, the graph of y œ g# (x) is the reflection of the graph of y œ g" (x) across the x-axis. Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Chapter 1 Practice Exercises 45. 46. The graph of f# (x) œ f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y œ f" (x), x 0 across the y-axis. Whenever g" (x) is positive, the graph of y œ g# (x) œ kg" (x)k is the same as the graph of y œ g" (x). When g" (x) is negative, the graph of y œ g# (x) is the reflection of the graph of y œ g" (x) across the x-axis. 47. 48. The graph of f# (x) œ f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y œ f" (x), x 0 across the y-axis. 49. (a) y œ gax
3b  (c) y œ ga
xb (e) y œ 5 † gaxb " # The graph of f# (x) œ f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y œ f" (x), x 0 across the y-axis. (b) y œ gˆx  3# ‰
2 (d) y œ
gaxb (f) y œ ga5xb 50. (a) Shift the graph of f right 5 units (b) Horizontally compress the graph of f by a factor of 4 (c) Horizontally compress the graph of f by a factor of 3 and a then reflect the graph about the y-axis (d) Horizontally compress the graph of f by a factor of 2 and then shift the graph left "# unit. (e) Horizontally stretch the graph of f by a factor of 3 and then shift the graph down 4 units. (f) Vertically stretch the graph of f by a factor of 3, then reflect the graph about the x-axis, and finally shift the graph up "4 unit. 51. Reflection of the grpah of y œ Èx about the x-axis followed by a horizontal compression by a factor of 1 2 then a shift left 2 units. Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 35 36 Chapter 1 Functions 52. Reflect the graph of y œ x about the x-axis, followed by a vertical compression of the graph by a factor of 3, then shift the graph up 1 unit. 53. Vertical compression of the graph of y œ 1 x2 by a factor of 2, then shift the graph up 1 unit. 54. Reflect the graph of y œ x1Î3 about the y-axis, then compress the graph horizontally by a factor of 5. 55. 56. period œ 1 57. period œ 41 58. period œ 2 period œ 4 Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Chapter 1 Practice Exercises 59. 60. period œ 21 period œ 21 1 3 61. (a) sin B œ sin œ b c œ b # Ê b œ 2 sin 1 3 œ 2Š È3 # ‹ œ È3. By the theorem of Pythagoras, a#  b# œ c# Ê a œ Èc#
b# œ È4
3 œ 1. 1 3 (b) sin B œ sin œ b c œ 2 c Ê cœ 2 sin 13 œ È23 œ Š ‹ # 4 È3 # . Thus, a œ Èc#
b# œ ÊŠ È43 ‹
(2)# œ É 34 œ 62. (a) sin A œ a c Ê a œ c sin A (b) tan A œ a b Ê a œ b tan A 63. (a) tan B œ b a Ê aœ (b) sin A œ a c Ê cœ 64. (a) sin A œ a c (c) sin A œ a c b tan B œ a sin A È c #
b # c 65. Let h œ height of vertical pole, and let b and c denote the distances of points B and C from the base of the pole, measured along the flatground, respectively. Then, tan 50° œ hc , tan 35° œ hb , and b
c œ 10. Thus, h œ c tan 50° and h œ b tan 35° œ (c  10) tan 35° Ê c tan 50° œ (c  10) tan 35° Ê c (tan 50°
tan 35°) œ 10 tan 35° tan 35° Ê c œ tan10 50°
tan 35° Ê h œ c tan 50° œ 10 tan 35° tan 50° tan 50°
tan 35° ¸ 16.98 m. 66. Let h œ height of balloon above ground. From the figure at the right, tan 40° œ ha , tan 70° œ hb , and a  b œ 2. Thus, h œ b tan 70° Ê h œ (2
a) tan 70° and h œ a tan 40° Ê (2
a) tan 70° œ a tan 40° Ê a(tan 40°  tan 70°) 70° œ 2 tan 70° Ê a œ tan 240°tantan 70° Ê h œ a tan 40° œ 2 tan 70° tan 40° tan 40°tan 70° ¸ 1.3 km. 67. (a) (b) The period appears to be 41. Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 2 È3 . 37 38 Chapter 1 Functions (c) f(x  41) œ sin (x  41)  cos ˆ x#41 ‰ œ sin (x  21)  cos ˆ x#  21‰ œ sin x  cos since the period of sine and cosine is 21. Thus, f(x) has period 41. x # 68. (a) (b) D œ (
_ß 0)  (!ß _); R œ [
1ß 1] (c) f is not periodic. For suppose f has period p. Then f ˆ #"1  kp‰ œ f ˆ #"1 ‰ œ sin 21 œ 0 for all integers k. Choose k so large that " #1  kp  " 1 Ê 0 " (1/21)kp  1. But then f ˆ #"1  kp‰ œ sin Š (1/#1")kp ‹  0 which is a contradiction. Thus f has no period, as claimed. CHAPTER 1 ADDITIONAL AND ADVANCED EXERCISES 1. There are (infinitely) many such function pairs. For example, f(x) œ 3x and g(x) œ 4x satisfy f(g(x)) œ f(4x) œ 3(4x) œ 12x œ 4(3x) œ g(3x) œ g(f(x)). 2. Yes, there are many such function pairs. For example, if g(x) œ (2x  3)$ and f(x) œ x"Î$ , then (f ‰ g)(x) œ f(g(x)) œ f a(2x  3)$ b œ a(2x  3)$ b "Î$ œ 2x  3. 3. If f is odd and defined at x, then f(
x) œ
f(x). Thus g(
x) œ f(
x)
2 œ
f(x)
2 whereas
g(x) œ
(f(x)
2) œ
f(x)  2. Then g cannot be odd because g(
x) œ
g(x) Ê
f(x)
2 œ
f(x)  2 Ê 4 œ 0, which is a contradiction. Also, g(x) is not even unless f(x) œ 0 for all x. On the other hand, if f is even, then g(x) œ f(x)
2 is also even: g(
x) œ f(
x)
2 œ f(x)
2 œ g(x). 4. If g is odd and g(0) is defined, then g(0) œ g(
0) œ
g(0). Therefore, 2g(0) œ 0 Ê g(0) œ 0. 5. For (xß y) in the 1st quadrant, kxk  kyk œ 1  x Í x  y œ 1  x Í y œ 1. For (xß y) in the 2nd quadrant, kxk  kyk œ x  1 Í
x  y œ x  1 Í y œ 2x  1. In the 3rd quadrant, kxk  kyk œ x  1 Í
x
y œ x  1 Í y œ
2x
1. In the 4th quadrant, kxk  kyk œ x  1 Í x  (
y) œ x  1 Í y œ
1. The graph is given at the right. Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Chapter 1 Additional and Advanced Exercises 6. We use reasoning similar to Exercise 5. (1) 1st quadrant: y  kyk œ x  kxk Í 2y œ 2x Í y œ x. (2) 2nd quadrant: y  kyk œ x  kxk Í 2y œ x  (
x) œ 0 Í y œ 0. (3) 3rd quadrant: y  kyk œ x  kxk Í y  (
y) œ x  (
x) Í 0 œ 0 Ê all points in the 3rd quadrant satisfy the equation. (4) 4th quadrant: y  kyk œ x  kxk Í y  (
y) œ 2x Í 0 œ x. Combining these results we have the graph given at the right: 7. (a) sin# x  cos# x œ 1 Ê sin# x œ 1
cos# x œ (1
cos x)(1  cos x) Ê (1
cos x) œ Ê 1
cos x sin x œ sin# x 1cos x sin x 1cos x (b) Using the definition of the tangent function and the double angle formulas, we have # tan ˆ x# ‰ œ sin# ˆ x# ‰ cos# ˆ #x ‰ œ " x ‹‹ cos Š2 Š # # "
cos Š2 Š #x ‹‹ # œ 1
cos x 1cos x . 8. The angles labeled # in the accompanying figure are equal since both angles subtend arc CD. Similarly, the two angles labeled ! are equal since they both subtend arc AB. Thus, triangles AED and BEC are similar which )
b implies a
b c œ 2a cos a c Ê (a
c)(a  c) œ b(2a cos )
b) Ê a#
c# œ 2ab cos )
b# Ê c# œ a#  b#
2ab cos ). 9. As in the proof of the law of sines of Section 1.3, Exercise 61, ah œ bc sin A œ ab sin C œ ac sin B Ê the area of ABC œ "# (base)(height) œ "# ah œ "# bc sin A œ "# ab sin C œ "# ac sin B. 10. As in Section 1.3, Exercise 61, (Area of ABC)# œ œ " 4 (base)# (height)# œ " 4 a# h # œ " 4 a# b# sin# C a# b# a"
cos# Cb . By the law of cosines, c# œ a#  b#
2ab cos C Ê cos C œ Thus, (area of ABC)# œ œ " 4 " 16 " 4 a# b# a"
cos# Cb œ # Š4a# b#
aa#  b#
c# b ‹ œ " 16 " 4 a# b# Œ"
Š a #  b#
c# ‹ #ab # œ a# b# 4 a#  b#
c# 2ab Š"
. aa #  b #
c # b 4a# b# # ‹ ca2ab  aa#  b#
c# bb a2ab
aa#  b#
c# bbd " ca(a  b)#
c# b ac#
(a
b)# bd œ 16 c((a  b)  c)((a  b)
c)(c  (a
b))(c
(a
b))d a  b  c
a  b  c a
b  c a  b
c œ ˆ # ‰ ˆ # ‰ ˆ # ‰ ˆ # ‰‘ œ s(s
a)(s
b)(s
c), where s œ a#bc . œ " 16 Therefore, the area of ABC equals Ès(s
a)(s
b)(s
c) . 11. If f is even and odd, then f(
x) œ
f(x) and f(
x) œ f(x) Ê f(x) œ
f(x) for all x in the domain of f. Thus 2f(x) œ 0 Ê f(x) œ 0. Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 39 40 Chapter 1 Functions f(
x)  f(
(
x)) œ f(x) #f(
x) œ E(x) Ê E # even function. Define O(x) œ f(x)
E(x) œ f(x)
f(x) #f(
x) œ f(x)
#f(
x) . Then O(
x) œ f(
x)
#f(
(
x)) œ f(
x)#
f(x) œ
Š f(x)
#f(
x) ‹ œ
O(x) Ê O is an odd function 12. (a) As suggested, let E(x) œ f(x)  f(
x) # Ê E(
x) œ is an Ê f(x) œ E(x)  O(x) is the sum of an even and an odd function. (b) Part (a) shows that f(x) œ E(x)  O(x) is the sum of an even and an odd function. If also f(x) œ E" (x)  O" (x), where E" is even and O" is odd, then f(x)
f(x) œ 0 œ aE" (x)  O" (x)b
(E(x)  O(x)). Thus, E(x)
E" (x) œ O" (x)
O(x) for all x in the domain of f (which is the same as the domain of E
E" and O
O" ). Now (E
E" )(
x) œ E(
x)
E" (
x) œ E(x)
E" (x) (since E and E" are even) œ (E
E" )(x) Ê E
E" is even. Likewise, (O"
O)(
x) œ O" (
x)
O(
x) œ
O" (x)
(
O(x)) (since O and O" are odd) œ
(O" (x)
O(x)) œ
(O"
O)(x) Ê O"
O is odd. Therefore, E
E" and O"
O are both even and odd so they must be zero at each x in the domain of f by Exercise 11. That is, E" œ E and O" œ O, so the decomposition of f found in part (a) is unique. 13. y œ ax#  bx  c œ a Šx#  ba x  b# 4a# ‹
b# 4a  c œ a ˆx  b ‰# 2a
b# 4a c (a) If a  0 the graph is a parabola that opens upward. Increasing a causes a vertical stretching and a shift of the vertex toward the y-axis and upward. If a  0 the graph is a parabola that opens downward. Decreasing a causes a vertical stretching and a shift of the vertex toward the y-axis and downward. (b) If a  0 the graph is a parabola that opens upward. If also b  0, then increasing b causes a shift of the graph downward to the left; if b  0, then decreasing b causes a shift of the graph downward and to the right. If a  0 the graph is a parabola that opens downward. If b  0, increasing b shifts the graph upward to the right. If b  0, decreasing b shifts the graph upward to the left. (c) Changing c (for fixed a and b) by ?c shifts the graph upward ?c units if ?c  0, and downward
?c units if ?c  0. 14. (a) If a  0, the graph rises to the right of the vertical line x œ
b and falls to the left. If a  0, the graph falls to the right of the line x œ
b and rises to the left. If a œ 0, the graph reduces to the horizontal line y œ c. As kak increases, the slope at any given point x œ x! increases in magnitude and the graph becomes steeper. As kak decreases, the slope at x! decreases in magnitude and the graph rises or falls more gradually. (b) Increasing b shifts the graph to the left; decreasing b shifts it to the right. (c) Increasing c shifts the graph upward; decreasing c shifts it downward. 15. Each of the triangles pictured has the same base b œ v?t œ v(1 sec). Moreover, the height of each triangle is the same value h. Thus "# (base)(height) œ " # bh œ A" œ A# œ A$ œ á . In conclusion, the object sweeps out equal areas in each one second interval. 16. (a) Using the midpoint formula, the coordinates of P are ˆ a# 0 ß b# 0 ‰ œ ˆ #a ß b# ‰ . Thus the slope of OP œ ?y ?x œ b/2 a/2 œ b a . Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Chapter 1 Additional and Advanced Exercises (b) The slope of AB œ b
0 0
a 41 œ
ba . The line segments AB and OP are perpendicular when the product # of their slopes is
" œ ˆ ba ‰ ˆ
ba ‰ œ
ba# . Thus, b# œ a# Ê a œ b (since both are positive). Therefore, AB is perpendicular to OP when a œ b. 17. From the figure we see that 0 Ÿ ) Ÿ cos ) œ and AB œ AD œ 1. From trigonometry we have the following: sin ) œ sin ) cos ) . œ CD, and tan ) œ œ We can see that: w " area ˜AEB  area sector DB  area ˜ADC Ê # aAEbaEBb  "# aADb2 )  "# aADbaCDb AE AB œ AE, tan ) œ 1 2 CD AD EB AE Ê "# sin ) cos )  "# a"b2 )  "# a"batan )b Ê "# sin ) cos )  "# )  " sin ) # cos ) 18. af‰gbaxb œ fagaxbb œ aacx  db  b œ acx  ad  b and ag‰f baxb œ gafaxbb œ caax  bb  d œ acx  cb  d Thus af‰gbaxb œ ag‰f baxb Ê acx  ad  b œ acx  bc  d Ê ad  b œ bc  d. Note that fadb œ ad  b and gabb œ cb  d, thus af‰gbaxb œ ag‰f baxb if fadb œ gabb. Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. EB AB œ EB, 42 Chapter 1 Functions NOTES: Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. CHAPTER 2 LIMITS AND CONTINUITY 2.1 RATES OF CHANGE AND TANGENTS TO CURVES 1. (a) ?f ?x œ f(3)
f(2) 3
# 2. (a) ?g ?x œ g(1)
g(
1) 1
(
1) 3. (a) ?h ?t œ h ˆ 341 ‰
h ˆ 14 ‰ 1 31 4
4 œ ?g ?t œ g(1)
g(0) 1
0 (2
1)
(2  1) 1
0 4. (a) 5. ?R ?) œ R(2)
R(0) 2
0 6. ?P ?) œ P(2)
P(1) 2
1 7. (a) ?y ?x œ ?y ?x œ ?y ?x œ ?y ?x œ ?y ?x œ ?y ?x œ ?y ?x œ œ œ 28
9 1 œ œ œ 1
1 2 œ 19 (b) ?f ?x œ f(1)
f(
") 1
(
1) œ 2
0 # œ1 œ0 (b) ?g ?x œ g(0)
g(
2) 0
(
2) œ 0
4 # œ
2 (b) ?h ?t œ h ˆ 1# ‰
h ˆ 16 ‰ 1
1 # 6 œ ?g ?t œ g(1)
g(
1) 1
(
1) œ
1
1 1 # È 8 1
È 1 # œ
14 œ
12 3
" # œ (8
16  10)
("
%  &) 1 ˆa2  h b2
3 ‰
ˆ 2 2
3 ‰ h œ (b) 0
È3 1 3 œ
3 È 3 1 (2
1)
(2
") #1 œ0 œ1 œ2
2œ0 4  4h  h2
3
1 h œ 4h  h2 h œ 4  h. As h Ä 0, 4  h Ä 4 Ê at Pa2, 1b the slope is 4. (b) y
1 œ 4ax
2b Ê y
1 œ 4x
8 Ê y œ 4x
7 8. (a) ˆ 5
a1  h b 2 ‰
ˆ 5
1 2 ‰ h œ 5
1
2h
h2
4 h œ
2h
h2 h œ
2
h. As h Ä 0,
2
h Ä
2 Ê at Pa1, 4b the slope is
2. (b) y
4 œ a
2bax
1b Ê y
4 œ
2x  2 Ê y œ
2x  6 9. (a) ˆa2  h b2
2 a 2  h b
3 ‰
ˆ 2 2
2 a 2 b
3 ‰ h œ 4  4h  h2
4
2h
3
a
3b h œ 2h  h2 h œ 2  h. As h Ä 0, 2  h Ä 2 Ê at Pa2,
3b the slope is 2. (b) y
a
3b œ 2ax
2b Ê y  3 œ 2x
4 Ê y œ 2x
7. 10. (a) ˆa1  h b2
4 a 1  h b ‰
ˆ 1 2
4 a 1 b ‰ h œ 1  2h  h2
4
4h
a
3b h œ h2
2h h œ h
2. As h Ä 0, h
2 Ä
2 Ê at Pa1,
3b the slope is
2. (b) y
a
3b œ a
2bax
1b Ê y  3 œ
2x  2 Ê y œ
2x
1. 11. (a) a2  h b 3
2 3 h œ 8  12h  4h2  h3
8 h œ 12h  4h2  h3 h œ 12  4h  h2 . As h Ä 0, 12  4h  h2 Ä 12, Ê at Pa2, 8b the slope is 12. (b) y
8 œ 12ax
2b Ê y
8 œ 12x
24 Ê y œ 12x
16. 12. (a) 2
a1  h b3
ˆ 2
1 3 ‰ h œ 2
1
3h
3h2
h3
1 h œ
3h
3h2
h3 h œ
3
3h
h2 . As h Ä 0,
3
3h
h2 Ä
3, Ê at Pa1, 1b the slope is
3. (b) y
1 œ a
3bax
1b Ê y
1 œ
3x  3 Ê y œ
3x  4. 13. (a) a1  hb3
12a1  hb
ˆ13
12a"b‰ h 2 œ 1  3h  3h2  h3
12
12h
a
11b h œ
9h  3h2  h3 h œ
9  3h  h2 . As h Ä 0,
9  3h  h Ä
9 Ê at Pa1,
11b the slope is
9. (b) y
a
11b œ a
9bax
1b Ê y  11 œ
9x  9 Ê y œ
9x
2. Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 44 Chapter 2 Limits and Continuity 14. (a) ?y ?x œ a2  h b 3
3 a 2  h b 2  4
ˆ 2 3
3 a 2 b 2  4 ‰ h 2 8  12h  6h2  h3
12
12h
3h2  4
0 h œ œ 3h2  h3 h œ 3h  h2 . As h Ä 0, 3h  h Ä 0 Ê at Pa2, 0b the slope is 0. (b) y
0 œ 0ax
2b Ê y œ 0. 15. (a) ?p ?t Slope of PQ œ Q 650
225 20
10 650
375 20
14 650
475 20
16.5 650
550 20
18 Q" (10ß 225) Q# (14ß 375) Q$ (16.5ß 475) Q% (18ß 550) œ 42.5 m/sec œ 45.83 m/sec œ 50.00 m/sec œ 50.00 m/sec (b) At t œ 20, the sportscar was traveling approximately 50 m/sec or 180 km/h. 16. (a) Slope of PQ œ Q Q" (5ß 20) Q# (7ß 39) Q$ (8.5ß 58) Q% (9.5ß 72) 80
20 10
5 80
39 10
7 80
58 10
8.5 80
72 10
9.5 ?p ?t œ 12 m/sec œ 13.7 m/sec œ 14.7 m/sec œ 16 m/sec (b) Approximately 16 m/sec 17. (a) (b) ?p ?t œ 174
62 2004
2002 œ 112 # œ 56 thousand dollars per year (c) The average rate of change from 2001 to 2002 is ??pt œ 62
27 20022
2001 œ 35 thousand dollars per year. 111
62 The average rate of change from 2002 to 2003 is ??pt œ 2003
2002 œ 49 thousand dollars per year. So, the rate at which profits were changing in 2002 is approximatley "# a35  49b œ 42 thousand dollars 18. (a) F(x) œ (x  2)/(x
2) x 1.2 F(x)
4.0 ?F ?x ?F ?x ?F ?x œ ?g ?x ?g ?x œ œ œ 1.1
3.4 1.01
3.04 1.001
3.004 1.0001
3.0004 1
3
4.0
(
3) œ
5.0; 1.2
1
3.04
(
3) œ
4.04; 1.01
1
3.!!!%
(
3) œ
4.!!!%; 1.0001
1 ?F ?x ?F ?x œ œ
3.4
(
3) œ
4.4; 1.1
1
3.004
(
3) œ
4.!!%; 1.001
1 È g(2)
g(1) œ #2

1" ¸ 0.414213 2
1 È1  h
" g(1  h)
g(1) (1  h)
1 œ h ?g ?x œ g(1.5)
g(1) 1.5
1 (b) The rate of change of F(x) at x œ 1 is
4. 19. (a) œ œ È1.5
" 0.5 ¸ 0.449489 (b) g(x) œ Èx 1h È1  h ŠÈ1  h
1‹ /h 1.1 1.04880 1.01 1.004987 1.001 1.0004998 1.0001 1.0000499 1.00001 1.000005 1.000001 1.0000005 0.4880 0.4987 0.4998 0.499 0.5 0.5 (c) The rate of change of g(x) at x œ 1 is 0.5. Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. per year. Section 2.1 Rates of Change and Tangents to Curves (d) The calculator gives lim hÄ! 20. (a) i) ii) (b) f(3)
f(2) 3
2 œ f(T)
f(2) T
# œ "
" 3 # 1 " " T
# T
# T f(T) af(T)
f(2)b/aT
2b " œ 6 1 È1  h
" h œ "# . œ
"6
#TT T
# 2 #T œ 45 œ 2
T #T(T
2) 2.1 0.476190
0.2381 œ 2
T
#T(2
T) 2.01 0.497512
0.2488 œ
#"T , T Á 2 2.001 0.499750
0.2500 2.0001 0.4999750
0.2500 2.00001 0.499997
0.2500 2.000001 0.499999
0.2500 (c) The table indicates the rate of change is
0.25 at t œ 2. " ‰ (d) lim ˆ
#T œ
4" TÄ# NOTE: Answers will vary in Exercises 21 and 22. s 15
0 ˜s 20
15 10 ˜s 30
20 21. (a) Ò0, 1Ó: ˜ ˜t œ 1
0 œ 15 mphà Ò1, 2.5Ó: ˜t œ 2.5
1 œ 3 mphà Ò2.5, 3.5Ó: ˜t œ 3.5
2.5 œ 10 mph (b) At Pˆ "# , 7.5‰: Since the portion of the graph from t œ 0 to t œ 1 is nearly linear, the instantaneous rate of change will be almost the same as the average rate of change, thus the instantaneous speed at t œ " # is 15
7.5 1
0.5 œ 15 mi/hr. At Pa2, 20b: Since the portion of the graph from t œ 2 to t œ 2.5 is nearly linear, the instantaneous rate of change will
20 be nearly the same as the average rate of change, thus v œ 20 2.5
2 œ 0 mi/hr. For values of t less than 2, we have Slope of PQ œ Q ?s ?t 15
20 1
2 œ 5 mi/hr 19
20 1.5
2 œ 2 mi/hr 19.9
20 1.9
2 œ 1 mi/hr Q" (1ß 15) Q# (1.5ß 19) Q$ (1.9ß 19.9) Thus, it appears that the instantaneous speed at t œ 2 is 0 mi/hr. At Pa3, 22b: s Q Slope of PQ œ ? ?t 35
22 4
3 30
22 3.5
3 23
22 3.1
3 Q" (4ß 35) Q# (3.5ß 30) Q$ (3.1ß 23) Slope of PQ œ Q œ 13 mi/hr Q" (2ß 20) œ 16 mi/hr Q# (2.5ß 20) œ 10 mi/hr Q$ (2.9ß 21.6) 20
22 2
3 œ 2 mi/hr 20
22 2.5
3 œ 4 mi/hr 21.6
22 2.9
3 œ 4 mi/hr Thus, it appears that the instantaneous speed at t œ 3 is about 7 mi/hr. (c) It appears that the curve is increasing the fastest at t œ 3.5. Thus for Pa3.5, 30b s Q Slope of PQ œ ? Q Slope of PQ œ ?t 35
30 4
3.5 œ 10 mi/hr 34
30 3.75
3.5 œ 16 mi/hr 32
30 3.6
3.5 œ 20 mi/hr Q" (4ß 35) Q# (3.75ß 34) Q$ (3.6ß 32) ˜A ˜t œ 10
15 3
0 (b) At Pa1, 14b: Q Q" (2ß 12.2) Q# (1.5ß 13.2) Q$ (1.1ß 13.85) ¸
1.67 gal day à Ò0, 5Ó: Q# (3.25ß 25) Q$ (3.4ß 28) Slope of PQ œ ˜A ˜t ?A ?t 12.2
14 2
1 œ
1.8 gal/day 13.2
14 1.5
1 œ
1.6 gal/day 13.85
14 1.1
1 œ
1.5 gal/day œ 3.9
15 5
0 ¸
2.2 gal day à Ò7, 10Ó: ?s ?t 22
30 3
3.5 œ 16 mi/hr 25
30 3.25
3.5 œ 20 mi/hr 28
30 3.4
3.5 œ 20 mi/hr Q" (3ß 22) Thus, it appears that the instantaneous speed at t œ 3.5 is about 20 mi/hr. 22. (a) Ò0, 3Ó: ?s ?t ˜A ˜t Q Q" (0ß 15) Q# (0.5ß 14.6) Q$ (0.9ß 14.86) œ 0
1.4 10
7 ¸
0.5 Slope of PQ œ gal day ?A ?t 15
14 0
1 œ
1 gal/day 14.6
14 0.5
1 œ
1.2 gal/day 14.86
14 0.9
1 œ
1.4 gal/day Thus, it appears that the instantaneous rate of consumption at t œ 1 is about
1.45 gal/day. At Pa4, 6b: Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 46 Chapter 2 Limits and Continuity Q Q" (5ß 3.9) Q# (4.5ß 4.8) Q$ (4.1ß 5.7) Slope of PQ œ 3.9
6 5
4 4.8
6 4.5
4 5.7
6 4.1
4 ?A ?t Q œ
2.1 gal/day Q" (3ß 10) œ
2.4 gal/day Q# (3.5ß 7.8) œ
3 gal/day Q$ (3.9ß 6.3) Slope of PQ œ 10
6 3
4 œ
4 gal/day 7.8
6 3.5
4 œ
3.6 gal/day 6.3
6 3.9
4 œ
3 gal/day Thus, it appears that the instantaneous rate of consumption at t œ 1 is
3 gal/day. At Pa8, 1b: Q Slope of PQ œ ??At Q Slope of PQ œ Q" (9ß 0.5) Q# (8.5ß 0.7) Q$ (8.1ß 0.95) 0.5
1 9
8 œ
0.5 gal/day 0.7
1 8.5
8 œ
0.6 gal/day 0.95
1 8.1
8 œ
0.5 gal/day Q" (7ß 1.4) 4.8
7.8 4.5
3.5 œ
3 gal/day 6
7.8 4
3.5 œ
3.6 gal/day 7.4
7.8 3.6
3.5 œ
4 gal/day Q" (2.5ß 11.2) Q# (7.5ß 1.3) Q$ (7.9ß 1.04) ?A ?t ?A ?t 1.4
1 7
8 œ
0.6 gal/day 1.3
1 7.5
8 œ
0.6 gal/day 1.04
1 7.9
8 œ
0.6 gal/day Thus, it appears that the instantaneous rate of consumption at t œ 1 is
0.55 gal/day. (c) It appears that the curve (the consumption) is decreasing the fastest at t œ 3.5. Thus for Pa3.5, 7.8b s Q Slope of PQ œ ??At Q Slope of PQ œ ? ?t Q" (4.5ß 4.8) Q# (4ß 6) Q$ (3.6ß 7.4) Q# (3ß 10) Q$ (3.4ß 8.2) Thus, it appears that the rate of consumption at t œ 3.5 is about
4 gal/day. 11.2
7.8 2.5
3.5 œ
3.4 gal/day 10
7.8 3
3.5 œ
4.4 gal/day 8.2
7.8 3.4
3.5 œ
4 gal/day 2.2 LIMIT OF A FUNCTION AND LIMIT LAWS 1. (a) Does not exist. As x approaches 1 from the right, g(x) approaches 0. As x approaches 1 from the left, g(x) approaches 1. There is no single number L that all the values g(x) get arbitrarily close to as x Ä 1. (b) 1 (c) 0 (d) 0.5 2. (a) 0 (b)
1 (c) Does not exist. As t approaches 0 from the left, f(t) approaches
1. As t approaches 0 from the right, f(t) approaches 1. There is no single number L that f(t) gets arbitrarily close to as t Ä 0. (d)
1 3. (a) True (d) False (g) True (b) True (e) False (c) False (f) True 4. (a) False (d) True (b) False (e) True (c) True 5. x lim x Ä 0 kx k x kx k does not exist because x kx k œ x x œ 1 if x  0 and approaches
1. As x approaches 0 from the right, x kx k x kxk œ x
x œ
1 if x  0. As x approaches 0 from the left, approaches 1. There is no single number L that all the function values get arbitrarily close to as x Ä 0. 6. As x approaches 1 from the left, the values of " x
1 become increasingly large and negative. As x approaches 1 from the right, the values become increasingly large and positive. There is no one number L that all the function values get arbitrarily close to as x Ä 1, so lim x
" 1 does not exist. xÄ1 Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Section 2.2 Limit of a Function and Limit Laws 47 7. Nothing can be said about f(x) because the existence of a limit as x Ä x! does not depend on how the function is defined at x! . In order for a limit to exist, f(x) must be arbitrarily close to a single real number L when x is close enough to x! . That is, the existence of a limit depends on the values of f(x) for x near x! , not on the definition of f(x) at x! itself. 8. Nothing can be said. In order for lim f(x) to exist, f(x) must close to a single value for x near 0 regardless of the xÄ0 value f(0) itself. 9. No, the definition does not require that f be defined at x œ 1 in order for a limiting value to exist there. If f(1) is defined, it can be any real number, so we can conclude nothing about f(1) from lim f(x) œ 5. xÄ1 10. No, because the existence of a limit depends on the values of f(x) when x is near 1, not on f(1) itself. If lim f(x) exists, its value may be some number other than f(1) œ 5. We can conclude nothing about lim f(x), xÄ1 xÄ1 whether it exists or what its value is if it does exist, from knowing the value of f(1) alone. 11. lim (2x  5) œ 2(
7)  5 œ
14  5 œ
9 x Ä
( 12. lim a
x#  5x
2b œ
(2)#  5(2)
2 œ
4  10
2 œ 4 xÄ# 13. lim 8(t
5)(t
7) œ 8(6
5)(6
7) œ
8 tÄ' 14. lim ax$
2x#  4x  8b œ (
2)$
2(
2)#  4(
2)  8 œ
8
8
8  8 œ
16 x Ä
# 15. lim x3 œ x Ä # x6 17. 23 26 16. lim# 3s(2s
1) œ 3 ˆ 23 ‰
2 ˆ 23 ‰
1‘ œ 2 ˆ 43
1‰ œ 5 8 sÄ $ lim 3(2x
1)# œ 3(2(
1)
1)# œ 3(
3)# œ 27 x Ä
" y2 18. lim # y Ä # y  5y  6 19. œ œ 22 (2)#  5(#)  6 œ 4 4  10  6 œ 4 #0 œ " 5 % lim (5
y)%Î$ œ [5
(
3)]%Î$ œ (8)%Î$ œ ˆ(8)"Î$ ‰ œ 2% œ 16 y Ä
$ 20. lim (2z
8)"Î$ œ (2(0)
8)"Î$ œ (
8)"Î$ œ
2 zÄ! 21. lim 3 22. lim È5h  4
2 h h Ä ! È3h  1  1 hÄ0 œ 5 È4  2 23. lim œ x
5 # x Ä & x
25 24. œ 3 È3(0)  1  1 œ lim hÄ0 œ 3 È1  1 È5h  4
2 h † œ 3 2 È5h  4  2 È5h  4  2 œ lim a5h  4b
4 h Ä 0 hŠÈ5h  4  2‹ œ lim 5h h Ä 0 hŠÈ5h  4  2‹ œ lim 5 4 x
5 œ lim x3 x Ä & (x  5)(x
5) lim # x Ä
$ x  4x  3 œ lim œ lim x3 1 x Ä & x5 x Ä
$ (x  3)(x  1) œ lim œ " 55 œ " 10 1 œ "
3  1 x Ä
$ x  1 5 h Ä 0 È5h  4  2 œ
"2 Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 2 3 48 25. Chapter 2 Limits and Continuity x#  3x
"0 x5 lim x Ä
& (x  5)(x
2) x5 œ lim x Ä
& (x
5)(x
2) x
2 26. lim x#
7x  "0 x
# œ lim 27. lim t#  t
2 t#
1 t Ä " (t
1)(t  1) xÄ# tÄ" t#  3t  2 # t Ä
" t
t
2 29. lim $ # x Ä
# x  2x lim
2x
4 5y$  8y# 1 x
1 x x Ä 1
1 32. lim xÄ0 33. lim 1 xc1 u%
" y# (5y  8) œ lim uÄ1 œ lim 4x
x# x Ä % 2
Èx œ lim 36. lim lim x Ä
" œ (v
2) av#  2v  4b Èx
3 x Ä * ˆÈ x
3 ‰ ˆ È x  3 ‰ x
1 x Ä 1 Èx  3
2 37. lim œ Èx#  12
4 x
2 xÄ2 x Ä
" œ lim xÄ2 (x
2)(x  2) x2 x Ä
2 È x #  5
3 lim x Ä
2 8
16 1 ‰ x
1 œ lim xÄ1 œ lim uÄ1 au#  "b (u  1) u#  u  1 " œ lim (x  2)(x
2) 444 (4)(8) œ x
1 x2 œ
2 4 3 12 32 œ 3 8 xÄ% œ lim ŠÈx  3  #‹ œ È4  2 œ 4 xÄ1 ax #  8 b
* œ lim x Ä
1 (x  1) ŠÈx#  8  $‹
2 33 œ (x
2) ŠÈx#  12  4‹ x Ä 2 Èx#  12  4 œ 2
1 œ lim x ˆ2  Èx‰ œ 4(2  2) œ 16 (x
1) ˆÈx  3  #‰ (x  3)
4 xÄ1 x Ä
1 È x #  )  $ œ œ " 6 œ lim (x  1) ŠÈx#  8  $‹ œ
"3 ax#  12b
16 œ lim x Ä 2 (x
2) ŠÈx#  12  4‹ œ 4 È16  4 ax  2b ŠÈx#  5  3‹ È x#  5  3 x
2 x Ä
2 œ lim " È9  3 œ x Ä * Èx  3 (1  1)(1  1) 111 œ # v Ä # (v  2) av  4b x ˆ2  È x ‰ ˆ 2
È x ‰ 2
Èx xÄ% œ lim œ v#  2v  4 œ lim œ lim œ lim 2 x Ä 1 ax
1bax  1b x Ä
2 ŠÈx#  5
3‹ ŠÈx#  5  3‹ ax  2b ŠÈx#  5  3‹ œ
#" œ lim
x1 œ
1 ŠÈx#  12
4‹ ŠÈx#  12  4‹ x Ä 2 (x
2) ŠÈx#  12  4‹ œ œ ŠÈx#  8
$‹ ŠÈx#  8  $‹ lim (x  1)(x
1) lim x Ä
1 (x  1) ŠÈx#  )  $‹ lim 5y  8 (x
1) ˆÈx  3  2‰ È ˆ x  3
#‰ ˆ È x  3  #‰ xÄ1 39. lim 40. œ
21 œ lim È x#  8
3 x1 œ lim
2 4 xÄ1 # v Ä # (v
2)(v  2) av  4b x(4
x) x Ä % 2
Èx œ
2 œ
13 1 œ lim Š ax
12x bax  1b † x ‹ œ lim au#  "b (u  1)(u
1) au#  u  1b (u
1) œ lim Èx
3 x
9 xÄ* x xÄ1 œ lim 35. lim b 1b b ax c 1b c 1bax b 1b ax 3 #
1  2
1
2 # y Ä ! 3y
16 xÄ1 œ œ # x Ä
# x œ lim ˆ 1
x x † ax œ lim t2 t Ä
" t
2 œ lim 1cx x 12 11 œ œ lim œ lim x Ä 1 x
1 % v Ä # v
16 t2
2(x  2) v$
8 34. lim xÄ# œ lim œ lim $ u Ä 1 u
1 38. t Ä
" (t
2)(t  1) # # y Ä ! y a3y
16b  x b1 1 x œ lim (x
5) œ 2
5 œ
3 t Ä " t1 (t  2)(t  1) œ lim x Ä
& œ lim # x Ä
# x (x  2) % # y Ä 0 3y
16y 31. lim (t  2)(t
1) œ lim 28. 30. lim xÄ# œ lim (x
2) œ
&
# œ
7 œ œ œ lim " 2 ax  2b ŠÈx#  5  3‹ x Ä
2 È9  3
4 ax #  5 b
9 œ
23 Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Section 2.2 Limit of a Function and Limit Laws 41. 2
È x#
5 x3 x Ä
3 lim œ Š2
Èx#
5‹ Š2  Èx#
5‹ œ lim (x  3) Š2  Èx#
5‹ x Ä
3 9
x# lim œ x Ä
3 (x  3) Š2  Èx#
5‹ 4
x x Ä 4 5
È x#  9 œ lim x Ä 4 Š5
Èx#  9‹ Š5  Èx#  9‹ a4
xb Š5  Èx#  9‹ xÄ4 x Ä
3 (x  3) Š2  Èx#
5‹ a4
xb Š5  Èx#  9‹ œ lim 42. lim 16
x# x Ä
3 (x  3) Š2  Èx#
5‹ (3
x)(3  x) lim œ lim (4
x)(4  x) xÄ4 œ lim 3
x x Ä
3 2  È x #
5 œ 6 2  È4 œ 3 2 a4
xb Š5  Èx#  9‹ 25
ax#  9b xÄ4 a4
xb Š5  Èx#  9‹ œ lim 4
ax #
5 b œ lim œ lim xÄ4 5  È x#  9 4x œ 5  È25 8 œ 5 4 2 43. lim a2sin x
1b œ 2sin 0
1 œ 0
1 œ
1 44. lim sin2 x œ Š lim sin x‹ œ asin 0b2 œ 02 œ 0 45. lim sec x œ 46. lim tan x œ xÄ0 xÄ0 47. lim xÄ0 1  x  sin x 3cos x 1 lim x Ä 0 cos x œ œ 1  0  sin 0 3cos 0 1 cos 0 œ œ 1 1 xÄ0 œ1 100 3 œ xÄ0 xÄ0 sin x lim x Ä 0 cos x œ sin 0 cos 0 œ 0 1 œ0 1 3 48. lim ax2
1ba2
cos xb œ a02
1ba2
cos 0b œ a
1ba2
1b œ a
1ba1b œ
1 xÄ0 49. x Ä lim
1Èx  4 cosax  1b œ x Ä lim
1Èx  4 † x Ä lim
1cosax  1b œ È
1  4 † cos 0 œ È4
1 † 1 œ È4
1 50. lim È7  sec2 x œ É lim a7  sec2 xb œ É7  lim sec2 x œ È7  sec2 0 œ É7  a1b2 œ 2È2 xÄ0 xÄ0 xÄ0 51. (a) quotient rule (c) sum and constant multiple rules (b) difference and power rules 52. (a) quotient rule (c) difference and constant multiple rules (b) power and product rules 53. (a) xlim f(x) g(x) œ ’xlim f(x)“ ’ x lim g(x)“ œ (5)(
2) œ
10 Äc Äc Äc (b) xlim 2f(x) g(x) œ 2 ’xlim f(x)“ ’ xlim g(x)“ œ 2(5)(
2) œ
20 Äc Äc Äc (c) xlim [f(x)  3g(x)] œ xlim f(x)  3 xlim g(x) œ 5  3(
2) œ
1 Äc Äc Äc lim f(x) f(x) 5 5 xÄc (d) xlim œ lim f(x)
lim g(x) œ 5
(
2) œ 7 Ä c f(x)
g(x) x 54. (a) (b) (c) (d) 55. (a) (b) Äc x Äc lim [g(x)  3] œ lim g(x)  lim 3 œ
$  $ œ ! xÄ% xÄ% xÄ% lim xf(x) œ lim x † lim f(x) œ (4)(0) œ 0 xÄ% xÄ% xÄ% # lim [g(x)] œ ’ lim g(x)“ œ [
3]# œ 9 xÄ% # g(x) x Ä % f(x)
1 lim xÄ% œ Ä% lim g(x) x lim f(x)
lim 1 xÄ% xÄ% œ
3 0
1 œ3 lim [f(x)  g(x)] œ lim f(x)  lim g(x) œ 7  (
3) œ 4 xÄb xÄb xÄb lim f(x) † g(x) œ ’ lim f(x)“ ’ lim g(x)“ œ (7)(
3) œ
21 xÄb xÄb xÄb Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 49 50 Chapter 2 Limits and Continuity lim 4g(x) œ ’ lim 4“ ’ lim g(x)“ œ (4)(
3) œ
12 (c) xÄb xÄb xÄb xÄb xÄb 7
3 œ
37 lim [p(x)  r(x)  s(x)] œ lim p(x)  lim r(x)  lim s(x) œ 4  0  (
3) œ 1 56. (a) x Ä
# x Ä
# x Ä
# x Ä
# lim p(x) † r(x) † s(x) œ ’ lim p(x)“ ’ lim r(x)“ ’ lim s(x)“ œ (4)(0)(
3) œ 0 (b) x Ä
# x Ä
# x Ä
# x Ä
# lim [
4p(x)  5r(x)]/s(x) œ ’
4 lim p(x)  5 lim r(x)“ ‚ lim s(x) œ [
4(4)  5(0)]/
3 œ (c) x Ä
# 57. lim hÄ! x Ä
# (1  h)#
1# h œ lim hÄ! (
2  h)#
(
2)# h hÄ! 58. lim 59. lim hÄ! ˆ #"
h ‰
ˆ "# ‰ h hÄ! È7  h
È7 h hÄ! 61. lim 1  2h  h#
1 h œ lim hÄ! 4
4hh#
4 h hÄ! œ lim [3(2  h)
4]
[3(2)
4] h 60. lim œ xÄb lim f(x)/g(x) œ lim f(x)/ lim g(x) œ (d) œ lim 2 2
h
" œ lim h(2  h) h œ lim hÄ! hÄ! œ lim (h
4) œ
4 hÄ!
2
(
2  h)
2h(
#  h) hÄ! h ŠÈ7  h  È7‹
h œ lim œ
"4 h Ä ! h(4
2h) œ lim (7  h)
7 h Ä ! h ŠÈ7  h  È7‹ œ lim h h Ä ! h ŠÈ7hÈ7‹ œ lim È3(0  h)  1
È3(0)  1 h hÄ! œ lim 3 h Ä ! È3h  1  1 œ œ lim ŠÈ3h  1
"‹ ŠÈ3h  1  "‹ h ŠÈ3h  1 "‹ hÄ! (3h  1)
" œ lim h Ä ! h ŠÈ3h  1  1 ‹ œ lim 3h h Ä ! h ŠÈ3h  1  "‹ 3 # 63. lim È5
2x# œ È5
2(0)# œ È5 and lim È5
x# œ È5
(0)# œ È5; by the sandwich theorem, xÄ! xÄ! lim f(x) œ È5 xÄ! 64. lim a2
x# b œ 2
0 œ 2 and lim 2 cos x œ 2(1) œ 2; by the sandwich theorem, lim g(x) œ 2 xÄ! 65. (a) xÄ! lim Š1
xÄ! x# 6‹ œ1
0 6 xÄ! œ 1 and lim 1 œ 1; by the sandwich theorem, lim (b) For x Á 0, y œ (x sin x)/(2
2 cos x) lies between the other two graphs in the figure, and the graphs converge as x Ä 0. 66. (a) lim Š "#
xÄ! lim xÄ! 1
cos x x# x# 24 ‹ œ lim 1 xÄ! #
lim x# x Ä ! #4 œ " # x sin x x Ä ! 2
2 cos x xÄ!
0œ " # and lim " xÄ! # œ1 œ "# ; by the sandwich theorem, œ "# . Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. " h Ä ! È 7 h  È 7 " #È 7 62. lim "6 3 œ lim (2  h) œ 2 h(h
4) h ŠÈ7  h
È7‹ ŠÈ7  h  È7‹ hÄ! x Ä
# œ3 œ lim
2h hÄ! œ lim 3h hÄ! h x Ä
# Section 2.2 Limit of a Function and Limit Laws (b) For all x Á 0, the graph of f(x) œ (1
cos x)/x# lies between the line y œ "# and the parabola yœ " #
x# /24, and the graphs converge as x Ä 0. 67. (a) f(x) œ ax#
*b/(x  3) x
3.1 f(x)
6.1
2.9
5.9 x f(x)
3.01
6.01
3.001
6.001
3.0001
6.0001
3.00001
6.00001
3.000001
6.000001
2.99
5.99
2.999
5.999
2.9999
5.9999
2.99999
5.99999
2.999999
5.999999 The estimate is lim f(x) œ
6. x Ä
$ (b) (c) f(x) œ x#
9 x3 œ (x  3)(x
3) x3 œ x
3 if x Á
3, and lim (x
3) œ
3
3 œ
6. x Ä
$ 68. (a) g(x) œ ax#
#b/ Šx
È2‹ x g(x) 1.4 2.81421 1.41 2.82421 1.414 2.82821 1.4142 2.828413 1.41421 2.828423 1.414213 2.828426 (b) (c) g(x) œ x#
2 x
È2 œ Šx  È2‹ Šx
È2‹ Šx
È2‹ œ x  È2 if x Á È2, and 69. (a) G(x) œ (x  6)/ ax#  4x
12b x
5.9
5.99 G(x)
.126582
.1251564 x G(x)
6.1
.123456
6.01
.124843
5.999
.1250156
6.001
.124984 lim x Ä È#
5.9999
.1250015
6.0001
.124998 Šx  È2‹ œ È2  È2 œ 2È2.
5.99999
.1250001
6.00001
.124999
5.999999
.1250000
6.000001
.124999 Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 51 52 Chapter 2 Limits and Continuity (b) (c) G(x) œ x6 ax#  4x
12b œ x6 (x  6)(x
2) œ " x
# 70. (a) h(x) œ ax#
2x
3b / ax#
4x  3b x 2.9 2.99 h(x) 2.052631 2.005025 x h(x) 3.1 1.952380 3.01 1.995024 " if x Á
6, and lim x Ä
' x
2 œ "
'
2 œ
"8 œ
0.125. 2.999 2.000500 2.9999 2.000050 2.99999 2.000005 2.999999 2.0000005 3.001 1.999500 3.0001 1.999950 3.00001 1.999995 3.000001 1.999999 (b) (c) h(x) œ x#
2x
3 x#
4x  3 œ (x
3)(x  1) (x
3)(x
1) œ x1 x
1 71. (a) f(x) œ ax#
1b / akxk
1b x
1.1
1.01 f(x) 2.1 2.01
.9 1.9 x f(x)
.99 1.99 if x Á 3, and lim x1 x Ä $ x
1 œ 31 3
1 œ 4 # œ 2.
1.001 2.001
1.0001 2.0001
1.00001 2.00001
1.000001 2.000001
.999 1.999
.9999 1.9999
.99999 1.99999
.999999 1.999999 (b) (c) f(x) œ x#
" kx k
1 (x  1)(x
1)
1 œ  (x  x1)(x
1)
(x  1) œ x  1, x 0 and x Á 1 , and lim (1
x) œ 1
(
1) œ 2. x Ä
1 œ 1
x, x  0 and x Á
1 Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Section 2.2 Limit of a Function and Limit Laws 72. (a) F(x) œ ax#  3x  2b / a2
kxkb x
2.1
2.01 F(x)
1.1
1.01
1.9
.9 x F(x)
1.99
.99
2.001
1.001
2.0001
1.0001
2.00001
1.00001
2.000001
1.000001
1.999
.999
1.9999
.9999
1.99999
.99999
1.999999
.999999 (b) (c) F(x) œ x#  3x  2 2
kx k (x  2)(x  1) œ  (x  2)(x#
x") 2x 73. (a) g()) œ (sin ))/) ) .1 g()) .998334 , x 0 , and lim (x  1) œ
2  1 œ
1. x Ä
# œ x  1, x  0 and x Á
2 .01 .999983 .001 .999999 .0001 .999999 .00001 .999999 .000001 .999999
.1 .998334
.01 .999983
.001 .999999
.0001 .999999
.00001 .999999
.000001 .999999 74. (a) G(t) œ (1
cos t)/t# t .1 G(t) .499583 .01 .499995 .001 .499999 .0001 .5 .00001 .5 .000001 .5
.1 .499583
.01 .499995
.001 .499999
.0001 .5
.00001 .5
.000001 .5 ) g()) lim g()) œ 1 )Ä! (b) t G(t) lim G(t) œ 0.5 tÄ! (b) Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 53 54 Chapter 2 Limits and Continuity 75. xlim f(x) exists at those points c where xlim x% œ xlim x# . Thus, c% œ c# Ê c# a1
c# b œ 0 Ê c œ 0, 1, or
1. Äc Äc Äc Moreover, lim f(x) œ lim x# œ 0 and lim f(x) œ lim f(x) œ 1. xÄ! xÄ! x Ä
1 xÄ1 76. Nothing can be concluded about the values of f, g, and h at x œ 2. Yes, f(2) could be 0. Since the conditions of the sandwich theorem are satisfied, lim f(x) œ
5 Á 0. xÄ# lim f(x)
lim 5 % xÄ% œ xÄlim x
lim 2 œ f(x)
5 x Ä % x
2 77. 1 œ lim xÄ% lim f(x)
5 xÄ% %
# xÄ% xÄ% 78. (a) 1 œ lim f(x) x# lim f(x) lim f(x) œ xÄlim# x# œ xÄ %# Ê (b) 1 œ lim f(x) x# œ ’ lim x Ä
# x Ä
# xÄ # x Ä
# 79. (a) 0 œ 3 † 0 œ ’ lim xÄ# Ê lim f(x)
5 œ 2(1) Ê lim f(x) œ 2  5 œ 7. xÄ% lim f(x) œ 4. x Ä
# f(x) lim x" “ x “ ’x Ä
# œ ’ lim x Ä
# f(x) ˆ " ‰ x “
# Ê lim x Ä
# f(x) x œ
2. f(x)
5 x
# “ ’xlim Ä#
5 (x
2)“ œ lim ’Š f(x) x
# ‹ (x
2)“ œ lim [f(x)
5] œ lim f(x)
5 f(x)
5 x
# “ ’xlim Ä# (x
2)“ Ê lim f(x) œ 5 as in part (a). xÄ# Ê lim f(x) œ 5. xÄ# xÄ# xÄ# (b) 0 œ 4 † 0 œ ’ lim xÄ# 80. (a) 0 œ 1 † 0 œ ’ lim f(x) # “ ’ lim xÄ! x xÄ! (b) 0 œ 1 † 0 œ 81. (a) lim x sin xÄ! (b)
1 Ÿ sin 82. (a) " x ’ lim f(x) # “ ’ lim xÄ! x xÄ! " x xÄ# # x“ œ ’ lim f(x) # xÄ! x x“ œ lim ’ f(x) x# xÄ! # “ ’ lim x# “ œ lim ’ f(x) x# † x “ œ lim f(x). That is, lim f(x) œ 0. xÄ! † x“ œ xÄ! lim f(x) . xÄ! x That is, xÄ! lim f(x) xÄ! x œ 0. œ0 Ÿ 1 for x Á 0: x  0 Ê
x Ÿ x sin " x Ÿ x Ê lim x sin " x œ 0 by the sandwich theorem; x  0 Ê
x " x x Ê lim x sin " x œ 0 by the sandwich theorem. x sin xÄ! xÄ! lim x# cos ˆ x"$ ‰ œ 0 xÄ! Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. xÄ! Section 2.3 The Precise Definition of a Limit (b)
1 Ÿ cos ˆ x"$ ‰ Ÿ 1 for x Á 0 Ê
x# Ÿ x# cos ˆ x"$ ‰ Ÿ x# Ê lim x# cos ˆ x"$ ‰ œ 0 by the sandwich xÄ! theorem since lim x# œ 0. xÄ! 83-88. Example CAS commands: Maple: f := x -> (x^4
16)/(x
2); x0 := 2; plot( f(x), x œ x0-1..x0+1, color œ black, title œ "Section 2.2, #83(a)" ); limit( f(x), x œ x0 ); In Exercise 85, note that the standard cube root, x^(1/3), is not defined for x<0 in many CASs. This can be overcome in Maple by entering the function as f := x -> (surd(x+1, 3)
1)/x. Mathematica: (assigned function and values for x0 and h may vary) Clear[f, x] f[x_]:=(x3
x2
5x
3)/(x  1)2 x0=
1; h = 0.1; Plot[f[x],{x, x0
h, x0  h}] Limit[f[x], x Ä x0] 2.3 THE PRECISE DEFINITION OF A LIMIT 1. Step 1: Step 2: kx
5k  $ Ê
$  x
5  $ Ê
$  5  x  $  5 $  5 œ 7 Ê $ œ 2, or
$  5 œ 1 Ê $ œ 4. The value of $ which assures kx
5k  $ Ê 1  x  7 is the smaller value, $ œ 2. Step 1: Step 2: kx
2k  $ Ê
$  x
2  $ Ê
$  #  x  $  2
$  2 œ 1 Ê $ œ 1, or $  2 œ 7 Ê $ œ 5. The value of $ which assures kx
2k  $ Ê 1  x  7 is the smaller value, $ œ 1. Step 1: Step 2: kx
(
3)k  $ Ê
$  x  $  $ Ê
$
3  x  $
3
$
3 œ
7# Ê $ œ "# , or $
$ œ
"# Ê $ œ 5# . 2. 3. The value of $ which assures kx
(
3)k  $ Ê
7#  x 
"# is the smaller value, $ œ "# . 4. Step 1: ¸x
ˆ
3# ‰¸  $ Ê
$  x  3 #  $ Ê
$
3 # x$
3 # Step 2: œ
7# Ê $ œ #, or $
3# œ
"# Ê $ œ 1. The value of $ which assures ¸x
ˆ
3# ‰¸  $ Ê
7#  x 
"# is the smaller value, $ œ ". Step 1: ¸x
"# ¸  $ Ê
$  x

$
3 # 5. " #  $ Ê
$  " # x$ " # Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 55 56 Chapter 2 Limits and Continuity Step 2: " or $  #" œ 47 Ê $ œ 14 . "¸ 4 ¸ The value of $ which assures x
#  $ Ê 9  x 
$  " # œ 4 9 Ê $œ " 18 , 4 7 is the smaller value, $ œ " 18 . 6. Step 1: Step 2: kx
3k  $ Ê
$  x
3  $ Ê
$  3  x  $  3
$  $ œ 2.7591 Ê $ œ 0.2409, or $  $ œ 3.2391 Ê $ œ 0.2391. The value of $ which assures kx
3k  $ Ê 2.7591  x  3.2391 is the smaller value, $ œ 0.2391. 7. Step 1: Step 2: kx
5k  $ Ê
$  x
5  $ Ê
$  5  x  $  5 From the graph,
$  5 œ 4.9 Ê $ œ 0.1, or $  5 œ 5.1 Ê $ œ 0.1; thus $ œ 0.1 in either case. 8. Step 1: Step 2: kx
(
3)k  $ Ê
$  x  3  $ Ê
$
3  x  $
3 From the graph,
$
3 œ
3.1 Ê $ œ 0.1, or $
3 œ
2.9 Ê $ œ 0.1; thus $ œ 0.1. 9. Step 1: Step 2: kx
1k  $ Ê
$  x
1  $ Ê
$  1  x  $  1 9 7 From the graph,
$  1 œ 16 Ê $ œ 16 , or $  1 œ 25 16 Ê $ œ 10. Step 1: Step 2: kx
3k  $ Ê
$  x
3  $ Ê
$  3  x  $  3 From the graph,
$  3 œ 2.61 Ê $ œ 0.39, or $  3 œ 3.41 Ê $ œ 0.41; thus $ œ 0.39. 11. Step 1: kx
2k  $ Ê
$  x
2  $ Ê
$  2  x  $  2 From the graph,
$  2 œ È3 Ê $ œ 2
È3 ¸ 0.2679, or $  2 œ È5 Ê $ œ È5
2 ¸ 0.2361; thus $ œ È5
2. Step 2: 12. Step 1: Step 2: 9 16 ; thus $ œ kx
(
1)k  $ Ê
$  x  1  $ Ê
$
1  x  $
1 From the graph,
$
1 œ
thus $ œ È5
2 # . È5 # Ê $œ È5
2 # ¸ 0.1180, or $
1 œ
13. Step 1: Step 2: kx
(
1)k  $ Ê
$  x  1  $ Ê
$
1  x  $
1 7 16 From the graph,
$
1 œ
16 9 Ê $ œ 9 ¸ 0.77, or $
1 œ
25 Ê 14. Step 1: ¸x
"# ¸  $ Ê
$  x
Step 2: 7 16 . From the graph,
$  thus $ œ 0.00248. " # œ " #  1 2.01 $ Ê
$  Ê $œ 1 2
" # x$ " #.01 " # ¸ 0.00248, or $  " # œ È3 # 9 25 Ê $œ 2
È3 # œ 0.36; thus $ œ 1 1.99 Ê $œ 1 1.99
¸ 0.1340; 9 25 " # œ 0.36. ¸ 0.00251; 15. Step 1: Step 2: k(x  1)
5k  0.01 Ê kx
4k  0.01 Ê
0.01  x
4  0.01 Ê 3.99  x  4.01 kx
4k  $ Ê
$  x
4  $ Ê
$  4  x  $  4 Ê $ œ 0.01. 16. Step 1: k(2x
2)
(
6)k  0.02 Ê k2x  4k  0.02 Ê
0.02  2x  4  0.02 Ê
4.02  2x 
3.98 Ê
2.01  x 
1.99 kx
(
2)k  $ Ê
$  x  2  $ Ê
$
2  x  $
2 Ê $ œ 0.01. Step 2: 17. Step 1: Step 2: ¹Èx  1
"¹  0.1 Ê
0.1  Èx  1
"  0.1 Ê 0.9  Èx  1  1.1 Ê 0.81  x  1  1.21 Ê
0.19  x  0.21 kx
0k  $ Ê
$  x  $ . Then,
$ œ
!Þ"* Ê $ œ !Þ"* or $ œ !Þ#"; thus, $ œ 0.19. Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Section 2.3 The Precise Definition of a Limit 18. Step 1: Step 2: ¸Èx
"# ¸  0.1 Ê
0.1  Èx
"#  0.1 Ê 0.4  Èx  0.6 Ê 0.16  x  0.36 ¸x
"4 ¸  $ Ê
$  x
4"  $ Ê
$  4"  x  $  4" . Then,
$  19. Step 1: Step 2: 20. Step 1: Step 2: 21. Step 1: Step 2: 22. Step 1: Step 2: 57 " 4 œ 0.16 Ê $ œ 0.09 or $  " 4 œ 0.36 Ê $ œ 0.11; thus $ œ 0.09. ¹È19
x
$¹  " Ê
"  È19
x
$  1 Ê 2  È19
x  % Ê 4  19
x  16 Ê
%  x
19 
16 Ê 15  x  3 or 3  x  15 kx
10k  $ Ê
$  x
10  $ Ê
$  10  x  $  10. Then
$  10 œ 3 Ê $ œ 7, or $  10 œ 15 Ê $ œ 5; thus $ œ 5. ¹Èx
7
4¹  1 Ê
"  Èx
7
%  1 Ê 3  Èx
7  5 Ê 9  x
7  25 Ê 16  x  32 kx
23k  $ Ê
$  x
23  $ Ê
$  23  x  $  23. Then
$  23 œ 16 Ê $ œ 7, or $  23 œ 32 Ê $ œ 9; thus $ œ 7. ¸ "x
4" ¸  0.05 Ê
0.05  " x
" 4  0.05 Ê 0.2  " x  0.3 Ê kx
4k  $ Ê
$  x
4  $ Ê
$  4  x  $  4. 2 2 Then
$  % œ 10 3 or $ œ 3 , or $  4 œ 5 or $ œ 1; thus $ œ 3 . 10 # x 10 3 or 10 3  x  5. kx#
3k  !.1 Ê
0.1  x#
3  0.1 Ê 2.9  x#  3.1 Ê È2.9  x  È3.1 ¹x
È3¹  $ Ê
$  x
È3  $ Ê
$  È3  x  $  È3. Then
$  È3 œ È2.9 Ê $ œ È3
È2.9 ¸ 0.0291, or $  È3 œ È3.1 Ê $ œ È3.1
È3 ¸ 0.0286; thus $ œ 0.0286. 23. Step 1: Step 2: kx#
4k  0.5 Ê
0.5  x#
4  0.5 Ê 3.5  x#  4.5 Ê È3.5  kxk  È4.5 Ê
È4.5  x 
È3.5, for x near
2. kx
(
2)k  $ Ê
$  x  2  $ Ê
$
#  x  $
2. Then
$
# œ
È4.5 Ê $ œ È4.5
# ¸ 0.1213, or $
# œ
È3.5 Ê $ œ #
È3.5 ¸ 0.1292; thus $ œ È4.5
2 ¸ 0.12. 24. Step 1: Step 2: 25. Step 1: Step 2: ¸ "x
(
1)¸  0.1 Ê
0.1  " x  1  0.1 Ê
11 10  " x 9 10 10 10 
10 Ê
10 11  x 
9 or
9  x 
11 . kx
(
1)k  $ Ê
$  x  1  $ Ê
$
"  x  $
". " 10 " Then
$
" œ
10 9 Ê $ œ 9 , or $
" œ
11 Ê $ œ 11 ; thus $ œ " 11 . kax#
5b
11k  " Ê kx#
16k  1 Ê
"  x#
16  1 Ê 15  x#  17 Ê È15  x  È17. kx
4k  $ Ê
$  x
4  $ Ê
$  %  x  $  %. Then
$  % œ È15 Ê $ œ %
È15 ¸ 0.1270, or $  % œ È17 Ê $ œ È17
% ¸ 0.1231; thus $ œ È17
4 ¸ 0.12. 26. Step 1: Step 2: 27. Step 1: Step 2: ¸ 120 ¸ x
5  " Ê
"  120 x
&1 Ê 4 120 x 6 Ê " 4  x 120  " 6 Ê 30  x  20 or 20  x  30. kx
24k  $ Ê
$  x
24  $ Ê
$  24  x  $  24. Then
$  24 œ 20 Ê $ œ 4, or $  24 œ 30 Ê $ œ 6; thus Ê $ œ 4. kmx
2mk  0.03 Ê
0.03  mx
2m  0.03 Ê
0.03  2m  mx  0.03  2m Ê 0.03 2
0.03 m x2 m . kx
2k  $ Ê
$  x
2  $ Ê
$  2  x  $  2. 0.03 0.03 Then
$  2 œ 2
0.03 m Ê $ œ m , or $  2 œ #  m Ê $ œ 0.03 m . In either case, $ œ Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 0.03 m . 58 Chapter 2 Limits and Continuity kmx
3mk  c Ê
c  mx
3m  c Ê
c  3m  mx  c  3m Ê 3
28. Step 1: kx
3k  $ Ê
$  x
3  $ Ê
$  $  B  $  $. Then
$  $ œ $
mc Ê $ œ mc , or $  $ œ $  mc Ê $ œ Step 2: ¸(mx  b)
ˆ m#  b‰¸  - Ê
c  mx
m#  c Ê
c  ¸x
"# ¸  $ Ê
$  x
"#  $ Ê
$  "#  x  $  "# . 29. Step 1: Step 2: Then
$  " # œ " #
c m Ê $œ c m, or $  " # œ " #  c m c m. m # Ê $œ c m  x 3 In either case, $ œ c m. c m. " #
In either case, $ œ c m.  mx  c  m # Ê c m c m x " #  c m. k(mx  b)
(m  b)k  0.05 Ê
0.05  mx
m  0.05 Ê
0.05  m  mx  0.05  m 0.05 Ê 1
0.05 m x" m . 30. Step 1: kx
1k  $ Ê
$  x
1  $ Ê
$  "  x  $  ". 0.05 0.05 Then
$  " œ "
0.05 m Ê $ œ m , or $  " œ "  m Ê $ œ Step 2: 0.05 m . In either case, $ œ 0.05 m . 31. lim (3
2x) œ 3
2(3) œ
3 xÄ3 ka3
2xb
(
3)k  0.02 Ê
0.02  6
2x  0.02 Ê
6.02 
2x 
5.98 Ê 3.01  x  2.99 or 2.99  x  3.01. 0  k x
3k  $ Ê
$  x
3  $ Ê
$  $  x  $  $ . Then
$  $ œ 2.99 Ê $ œ 0.01, or $  $ œ 3.01 Ê $ œ 0.01; thus $ œ 0.01. Step 1: Step 2: 32. lim (
3x
#) œ (
3)(
1)
2 œ 1 x Ä
1 k(
3x
2)
1k  0.03 Ê
0.03 
3x
3  0.03 Ê 0.01  x  1 
0.01 Ê
1.01  x 
0.99. kx
(
1)k  $ Ê
$  x  1  $ Ê
$
"  x  $
1. Then
$
" œ
1.01 Ê $ œ 0.01, or $
" œ
0.99 Ê $ œ 0.01; thus $ œ 0.01. Step 1: Step 2: 33. lim x#
4 x Ä # x
# 34. 35. œ lim xÄ# # (x  2)(x
2) (x
2) œ lim (x  2) œ #  # œ 4, x Á 2 xÄ# (x  2)(x
2) (x
2) Step 1: ¹Š xx

24 ‹ Step 2: Ê 1.95  x  2.05, x Á 2. kx
2k  $ Ê
$  x
2  $ Ê
$  2  x  $  2. Then
$  2 œ 1.95 Ê $ œ 0.05, or $  2 œ 2.05 Ê $ œ 0.05; thus $ œ 0.05. lim x Ä
& x#  6x  5 x5
4¹  0.05 Ê
0.05  œ lim x Ä
& (x  5)(x  1) (x  5)
%  0.05 Ê 3.95  x  2  4.05, x Á 2 œ lim (x  1) œ
4, x Á
5. x Ä
& (x  5)(x  ") (x  5) Step 1: # ¹Š x x 6x5 5 ‹ Step 2: Ê
5.05  x 
4.95, x Á
5. kx
(
5)k  $ Ê
$  x  5  $ Ê
$
&  x  $
&. Then
$
& œ
5.05 Ê $ œ 0.05, or $
& œ
4.95 Ê $ œ 0.05; thus $ œ 0.05.
(
4)¹  0.05 Ê
0.05   4  0.05 Ê
4.05  x  1 
3.95, x Á
5 lim È1
5x œ È1
5(
3) œ È16 œ 4 x Ä
$ Step 1: ¹È1
5x
4¹  0.5 Ê
0.5  È1
5x
4  0.5 Ê 3.5  È1
5x  4.5 Ê 12.25  1
5x  20.25 Step 2: Ê 11.25 
5x  19.25 Ê
3.85  x 
2.25. kx
(
3)k  $ Ê
$  x  3  $ Ê
$
$  x  $
$. Then
$
$ œ
3.85 Ê $ œ 0.85, or $
$ œ
2.25 Ê 0.75; thus $ œ 0.75. 36. lim 4 xÄ# x Step 1: œ 4 # œ2 ¸ 4x
2¸  0.4 Ê
0.4  4 x
2  0.4 Ê 1.6  4 x  2.4 Ê 10 16  x 4  10 24 Ê 10 4 x Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 10 6 or 5 3  x  25 . Section 2.3 The Precise Definition of a Limit Step 2: 59 kx
2k  $ Ê
$  x
2  $ Ê
$  #  x  $  #. Then
$  # œ 53 Ê $ œ "3 , or $  # œ #5 Ê $ œ "# ; thus $ œ 3" . 37. Step 1: Step 2: k(9
x)
5k  % Ê
%  4
x  % Ê
%
4 
x  %
4 Ê %  %  x  4
% Ê %
%  x  4  %. kx
4k  $ Ê
$  x
4  $ Ê
$  %  x  $  %. Then
$  4 œ
%  4 Ê $ œ %, or $  % œ %  % Ê $ œ %. Thus choose $ œ %. 38. Step 1: k(3x
7)
2k  % Ê
%  3x
9  % Ê 9
%  3x  *  % Ê 3
Step 2: 39. Step 1: Step 2: 40. Step 1: Step 2: 41. Step 1: Step 2: 42. Step 1: Step 2: 43. Step 1: Step 2:  x  3  3% . kx
3k  $ Ê
$  x
3  $ Ê
$  3  x  $  3. Then
$  3 œ $
3% Ê $ œ 3% , or $  3 œ 3  3% Ê $ œ 3% . Thus choose $ œ 3% . ¹Èx
5
2¹  % Ê
%  Èx
5
2  % Ê 2
%  Èx
5  2  % Ê (2
%)#  x
5  (2  %)# Ê (2
%)#  &  x  (2  %)#  5. kx
9k  $ Ê
$  x
9  $ Ê
$  9  x  $  9. Then
$  * œ %#
%%  * Ê $ œ %%
%# , or $  * œ %#  %%  * Ê $ œ %%  %# . Thus choose the smaller distance, $ œ %%
%# . ¹È4
x
2¹  % Ê
%  È4
x
2  % Ê 2
%  È4
x  2  % Ê (2
%)#  %
x  (2  %)# Ê
(2  %)#  x
4 
(2
%)# Ê
(2  %)#  %  x 
(2
%)#  %. k x
0k  $ Ê
$  x  $ . Then
$ œ
(2  %)#  4 œ
%#
%% Ê $ œ %%  %# , or $ œ
(2
%)#  4 œ 4%
%# . Thus choose the smaller distance, $ œ 4%
%# . For x Á 1, kx#
1k  % Ê
%  x#
"  % Ê "
%  x#  "  % Ê È1
%  kxk  È1  % Ê È"
%  x  È1  % near B œ ". kx
1k  $ Ê
$  x
1  $ Ê
$  "  x  $  ". Then
$  " œ È1
% Ê $ œ "
È1
%, or $  1 œ È"  % Ê $ œ È"  %
1. Choose $ œ min š"
È1
%ß È1  %
"›, that is, the smaller of the two distances. For x Á
2, kx#
4k  % Ê
%  x#
4  % Ê 4
%  x#  4  % Ê È4
%  kxk  È4  % Ê
È4  %  x 
È4
% near B œ
2. kx
(
2)k  $ Ê
$  x  2  $ Ê
$
2  x  $
2. Then
$
2 œ
È%  % Ê $ œ È%  %
#, or $
# œ
È%
% Ê $ œ #
È%
%. Choose $ œ min šÈ%  %
#ß #
È%
%› . ¸ "x
1¸  % Ê
%  " x
"% Ê "
%  " x "% Ê " 1% % "%, " 1
%. x kx
1k  $ Ê
$  x
1  $ Ê "
$  x  "  $ . Then "
$ œ " " % Ê $ œ "
" " % œ " % % , or "  $ œ "
" % Ê $ œ Choose $ œ 44. Step 1: % 3 " "
%
"œ % "
%. the smaller of the two distances. ¸ x"#
"3 ¸  % Ê
%  " x#
" 3 % Ê " 3
%  " x#  " 3 % Ê 1
3% 3  " x#  1  $% 3 3 È $. Ê É 1 3 $%  kxk  É "
3 $% , or É " 3 $%  x  É "
$ % for x near Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Ê 3 "
$%  x#  3 "  $% 60 Chapter 2 Limits and Continuity Step 2: ¹x
È3¹  $ Ê
$  x
È3  $ Ê È3
$  x  È3  $ . Then È3
$ œ É " 3 $% Ê $ œ È3
É " 3 $% , or È3  $ œ É "
3 $% Ê $ œ É "
3 $%
È3. Choose $ œ min šÈ3
É " 3 $% ß É "
3 $%
È3›. 45. Step 1: Step 2: 46. Step 1: Step 2: 47. Step 1: # ¹Š xx 
3* ‹
(
6)¹  % Ê
%  (x
3)  6  %, x Á
3 Ê
%  x  3  % Ê
%
$  x  %
$. kx
(
3)k  $ Ê
$  x  3  $ Ê
$
$  x  $
3. Then
$
$ œ
%
$ Ê $ œ %, or $
$ œ %
$ Ê $ œ %. Choose $ œ %. # ¹Š xx

11 ‹
2¹  % Ê
%  (x  1)
2  %, x Á 1 Ê "
%  x  "  %. kx
1k  $ Ê
$  x
1  $ Ê "
$  x  "  $ . Then "
$ œ "
% Ê $ œ %, or "  $ œ "  % Ê $ œ %. Choose $ œ %. x  1: l(4
2x)
2l  % Ê !  2
2x  % since x  1Þ Thus, 1
x Step 2: 48. Step 1: Step 2: % #  x  !; 1. Thus, " Ÿ x  1  6% . 1: l(6x
4)
2l  % Ê ! Ÿ 6x
6  % since x kx
1k  $ Ê
$  x
1  $ Ê "
$  x  1  $ . Then 1
$ œ "
#% Ê $ œ #% , or "  $ œ 1  6% Ê $ œ 6% . Choose $ œ 6% . x  !: k2x
0k  % Ê
%  2x  ! Ê
#%  x  0; x 0: ¸ x#
!¸  % Ê ! Ÿ x  #%. k x
0k  $ Ê
$  x  $ . Then
$ œ
#% Ê $ œ #% , or $ œ #% Ê $ œ #%. Choose $ œ #% . 49. By the figure,
x Ÿ x sin " x Ÿ x for all x  0 and
x x sin then by the sandwich theorem, in either case, lim x sin xÄ! 50. By the figure,
x# Ÿ x# sin " x " x " x x for x  0. Since lim (
x) œ lim x œ 0, xÄ! œ 0. xÄ! Ÿ x# for all x except possibly at x œ 0. Since lim a
x# b œ lim x# œ 0, then by the sandwich theorem, lim x# sin xÄ! " x xÄ! œ 0. xÄ! 51. As x approaches the value 0, the values of g(x) approach k. Thus for every number %  0, there exists a $  ! such that !  kx
0k  $ Ê kg(x)
kk  %. 52. Write x œ h  c. Then !  lx
cl  $ Í
$  x
c  $ , x Á c Í
$  ah  cb
c  $ , h  c Á c Í
$  h  $ , h Á ! Í !  lh
!l  $ . Thus, limfaxb œ L Í for any %  !, there exists $  ! such that lfaxb
Ll  % whenever !  lx
cl  $ x Äc Í lfah  cb
Ll  % whenever !  lh
!l  $ Í lim fah  cb œ L. hÄ! 53. Let f(x) œ x# . The function values do get closer to
1 as x approaches 0, but lim f(x) œ 0, not
1. The function f(x) œ x# never gets arbitrarily close to
1 for x near 0. xÄ! Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Section 2.3 The Precise Definition of a Limit 61 54. Let f(x) œ sin x, L œ "# , and x! œ 0. There exists a value of x (namely, x œ 16 ) for which ¸sin x
"# ¸  % for any given %  0. However, lim sin x œ 0, not "# . The wrong statement does not require x to be arbitrarily close to x! . As another\ xÄ! example, let g(x) œ sin "x , L œ #" , and x! œ 0. We can choose infinitely many values of x near 0 such that sin you can see from the accompanying figure. However, lim sin xÄ! " x " x œ " # as fails to exist. The wrong statement does not require all values of x arbitrarily close to x! œ 0 to lie within %  0 of L œ "# . Again you can see from the figure that there are also infinitely many values of x near 0 such that sin "x œ 0. If we choose %  ¸sin "x
#" ¸  % for all values of x sufficiently near x! œ 0. # 55. kA
*k Ÿ 0.01 Ê
0.01 Ÿ 1 ˆ x# ‰
9 Ÿ 0.01 Ê 8.99 Ÿ Ê 2É 8.99 1 ŸxŸ 2É 9.01 1 1 x# 4 " 4 we cannot satisfy the inequality Ÿ 9.01 Ê 4 1 (8.99) Ÿ x# Ÿ 4 1 (9.01) or 3.384 Ÿ x Ÿ 3.387. To be safe, the left endpoint was rounded up and the right endpoint was rounded down. 56. V œ RI Ê (120)(10) 51 V R ŸRŸ œ I Ê ¸ VR
5¸ Ÿ 0.1 Ê
0.1 Ÿ (120)(10) 49 120 R
5 Ÿ 0.1 Ê 4.9 Ÿ 120 R Ÿ 5.1 Ê 10 49 R 1#0 10 51 Ê Ê 23.53 Ÿ R Ÿ 24.48. To be safe, the left endpoint was rounded up and the right endpoint was rounded down. 57. (a)
$  x
1  0 Ê "
$  x  1 Ê f(x) œ x. Then kf(x)
2k œ kx
2k œ 2
x  2
1 œ 1. That is, kf(x)
2k 1 "# no matter how small $ is taken when "
$  x  1 Ê lim f(x) Á 2. xÄ1 (b) 0  x
1  $ Ê "  x  "  $ Ê f(x) œ x  1. Then kf(x)
1k œ k(x  1)
1k œ kxk œ x  1. That is, kf(x)
1k 1 no matter how small $ is taken when "  x  "  $ Ê lim f(x) Á 1. xÄ1 (c)
$  x
1  ! Ê "
$  x  1 Ê f(x) œ x. Then kf(x)
1.5k œ kx
1.5k œ 1.5
x  1.5
1 œ 0.5. Also, !  x
1  $ Ê 1  x  "  $ Ê f(x) œ x  1. Then kf(x)
1.5k œ k(x  1)
1.5k œ kx
0.5k œ x
0.5  "
0.5 œ 0.5. Thus, no matter how small $ is taken, there exists a value of x such that
$  x
1  $ but kf(x)
1.5k "# Ê lim f(x) Á 1.5. xÄ1 58. (a) For 2  x  2  $ Ê h(x) œ 2 Ê kh(x)
4k œ 2. Thus for %  2, kh(x)
4k matter how small we choose $  0 Ê lim h(x) Á 4. % whenever 2  x  2  $ no (b) For 2  x  2  $ Ê h(x) œ 2 Ê kh(x)
3k œ 1. Thus for %  1, kh(x)
3k matter how small we choose $  0 Ê lim h(x) Á 3. % whenever 2  x  2  $ no xÄ# xÄ# (c) For 2
$  x  2 Ê h(x) œ x# so kh(x)
2k œ kx#
2k . No matter how small $  0 is chosen, x# is close to 4 when x is near 2 and to the left on the real line Ê kx#
2k will be close to 2. Thus if %  1, kh(x)
2k % whenever 2
$  x  2 no mater how small we choose $  0 Ê lim h(x) Á 2. xÄ# Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 62 Chapter 2 Limits and Continuity 59. (a) For 3
$  x  3 Ê f(x)  4.8 Ê kf(x)
4k 0.8. Thus for %  0.8, kf(x)
4k 3
$  x  3 no matter how small we choose $  0 Ê lim f(x) Á 4. xÄ$ % whenever (b) For 3  x  3  $ Ê f(x)  3 Ê kf(x)
4.8k 1.8. Thus for %  1.8, kf(x)
4.8k no matter how small we choose $  0 Ê lim f(x) Á 4.8. % whenever 3  x  3  $ (c) For 3
$  x  3 Ê f(x)  4.8 Ê kf(x)
3k 1.8. Again, for %  1.8, kf(x)
3k no matter how small we choose $  0 Ê lim f(x) Á 3. % whenever $
$  x  3 xÄ$ xÄ$ 60. (a) No matter how small we choose $  0, for x near
1 satisfying
"
$  x 
"  $ , the values of g(x) are near 1 Ê kg(x)
2k is near 1. Then, for % œ "# we have kg(x)
2k "# for some x satisfying
"
$  x 
"  $ , or !  kx  1k  $ Ê lim g(x) Á 2. x Ä
1 (b) Yes, lim g(x) œ 1 because from the graph we can find a $  ! such that kg(x)
1k  % if !  kx
(
1)k  $ . x Ä
1 61-66. Example CAS commands (values of del may vary for a specified eps): Maple: f := x -> (x^4-81)/(x-3);x0 := 3; plot( f(x), x=x0-1..x0+1, color=black, # (a) title="Section 2.3, #61(a)" ); L := limit( f(x), x=x0 ); # (b) epsilon := 0.2; # (c) plot( [f(x),L-epsilon,L+epsilon], x=x0-0.01..x0+0.01, color=black, linestyle=[1,3,3], title="Section 2.3, #61(c)" ); q := fsolve( abs( f(x)-L ) = epsilon, x=x0-1..x0+1 ); # (d) delta := abs(x0-q); plot( [f(x),L-epsilon,L+epsilon], x=x0-delta..x0+delta, color=black, title="Section 2.3, #61(d)" ); for eps in [0.1, 0.005, 0.001 ] do # (e) q := fsolve( abs( f(x)-L ) = eps, x=x0-1..x0+1 ); delta := abs(x0-q); head := sprintf("Section 2.3, #61(e)\n epsilon = %5f, delta = %5f\n", eps, delta ); print(plot( [f(x),L-eps,L+eps], x=x0-delta..x0+delta, color=black, linestyle=[1,3,3], title=head )); end do: Mathematica (assigned function and values for x0, eps and del may vary): Clear[f, x] y1: œ L
eps; y2: œ L  eps; x0 œ 1; f[x_]: œ (3x2
(7x  1)Sqrt[x]  5)/(x
1) Plot[f[x], {x, x0
0.2, x0  0.2}] L: œ Limit[f[x], x Ä x0] eps œ 0.1; del œ 0.2; Plot[{f[x], y1, y2},{x, x0
del, x0  del}, PlotRange Ä {L
2eps, L  2eps}] Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Section 2.4 One-Sided Limits 2.4 ONE-SIDED LIMITS 1. (a) True (e) True (i) False (b) True (f) True (j) False (c) False (g) False (k) True (d) True (h) False (l) False 2. (a) True (e) True (i) True (b) False (f) True (j) False (c) False (g) True (k) True (d) True (h) True 3. (a) lim f(x) œ x Ä #b 2 #  " œ #, lim c f(x) œ $
# œ " xÄ# (b) No, lim f(x) does not exist because lim b f(x) Á lim c f(x) xÄ# xÄ# xÄ# (c) lim c f(x) œ 4#  1 œ 3, lim b f(x) œ 4#  " œ $ xÄ% xÄ% (d) Yes, lim f(x) œ 3 because 3 œ lim c f(x) œ lim b f(x) xÄ% xÄ% xÄ% 4. (a) lim f(x) œ x Ä #b 2 # œ 1, lim c f(x) œ $
# œ ", f(2) œ 2 xÄ# (b) Yes, lim f(x) œ 1 because " œ lim b f(x) œ lim c f(x) xÄ# xÄ# xÄ# (c) lim c f(x) œ 3
(
1) œ 4, lim b f(x) œ 3
(
1) œ 4 x Ä
" x Ä
" (d) Yes, lim f(x) œ 4 because 4 œ x Ä
" lim x Ä
"c f(x) œ lim x Ä
"b f(x) 5. (a) No, lim b f(x) does not exist since sin ˆ "x ‰ does not approach any single value as x approaches 0 xÄ! (b) lim c f(x) œ lim c 0 œ 0 xÄ! (c) xÄ! lim f(x) does not exist because lim b f(x) does not exist xÄ! xÄ! 6. (a) Yes, lim b g(x) œ 0 by the sandwich theorem since
Èx Ÿ g(x) Ÿ Èx when x  0 xÄ! (b) No, lim c g(x) does not exist since Èx is not defined for x  0 xÄ! (c) No, lim g(x) does not exist since lim c g(x) does not exist xÄ! xÄ! 7. (aÑ lim f(x) œ " œ lim b f(x) xÄ1 (c) Yes, lim f(x) œ 1 since the right-hand and left-hand (b) x Ä 1c xÄ1 limits exist and equal 1 8. (a) (b) lim f(x) œ 0 œ lim c f(x) xÄ1 x Ä 1b (c) Yes, lim f(x) œ 0 since the right-hand and left-hand xÄ1 limits exist and equal 0 Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 63 64 Chapter 2 Limits and Continuity 9. (a) domain: 0 Ÿ x Ÿ 2 range: 0  y Ÿ 1 and y œ 2 (b) xlim f(x) exists for c belonging to Äc (0ß 1)  ("ß #) (c) x œ 2 (d) x œ 0 10. (a) domain:
_  x  _ range:
" Ÿ y Ÿ 1 (b) xlim f(x) exists for c belonging to Äc (
_ß
1)  (
"ß ")  ("ß _) (c) none (d) none 11. x Ä
!Þ&c lim 13. x Ä
#b 14. x Ä 1c 15. h Ä !b lim 2 0.5  2 È3 É 3/2 É xx  É

1 œ
0.5  1 œ 1/2 œ lim x Ä 1b " 1 È0 œ ! É "1
É xx
# œ # œ 5‰ ˆ x x 1 ‰ ˆ 2x ˆ
2 ‰ 2(
2)  5 ˆ1‰ x#  x œ
2  1 Š (
2)#  (
2) ‹ œ (2) 2 œ 1 lim ˆ x " 1 ‰ ˆ x x 6 ‰ ˆ 3
7 x ‰ œ ˆ 1 " 1 ‰ ˆ 1 1 6 ‰ ˆ 3
7 1 ‰ œ ˆ "# ‰ ˆ 71 ‰ ˆ 27 ‰ œ 1 lim Èh#  4h  5
È5 h œ lim b hÄ! 16. 12. lim h Ä !c (b) 18. (a) (b) ah#  4h  5b
5 h ŠÈh#  4h  5  È5‹ È6
È5h#  11h  6 h œ lim c hÄ! 17. (a) œ lim b Š hÄ! x Ä
#c lim x Ä 1b lim x Ä 1c œ lim c Š hÄ! 6
a5h#  11h  6b lim lim œ lim b hÄ! h ŠÈ6  È5h#  11h  6‹ x Ä
#b kx 2 k x 2 (x  3) œ kx  2 k x2 œ lim lim x Ä
#c È2x (x
1) kx
1 k È2x (x
1) kx
1 k œ lim b xÄ1
h(5h  11) h ŠÈ6  È5h#  11h  6‹ (x  3) x Ä
#b œ œ 04 È5  È5 œ 2 È5 È5h#  11h  6 È6
È5h#  11h  6 È ‹ Š È66  ‹ h  È5h#  11h  6 x Ä
#b lim h(h  4) h ŠÈh#  4h  5  È5‹ œ lim c hÄ! œ (x  3) Èh#  4h  5
È5 È # 4h  5  È5 ‹ Š Èhh#  ‹ h  4h  5  È5 (x2) (x#)
(0  11) È6  È6 11 œ
2È 6 akx  2k œ ax  2b for x 
2b (x  3) œ a(
2)  3b œ 1 (x  3) ’
(x(x#2) ) “ lim œ x Ä
#c akx  2k œ
(x  2) for x 
2b (x  3)(
1) œ
(
2  3) œ
1 È2x (x
1) (x
1) akx
1k œ x
1 for x  1b œ lim b È2x œ È2 xÄ1 œ lim c xÄ1 È2x (x
1)
(x
1) akx
1k œ
(x
1) for x  1b œ lim c
È2x œ
È2 xÄ1 Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Section 2.4 One-Sided Limits 19. (a) ) Ä $b 20. (a) t Ä %b Ú) Û ) lim œ1 3 3 lim at
ÚtÛb œ 4
4 œ 0 21. lim sin È2) È 2) 22. lim sin kt t 23. lim sin 3y 4y )Ä! tÄ! yÄ! 24. œ tan 2x x xÄ! 25. lim 2t 27. lim xÄ! )Ä! œ lim c ˆ "3 † hÄ! sin 2x ‰ ˆ cos 2x x xÄ! œ lim œ 2 lim t sin t t Ä ! ˆ cos t ‰ x csc 2x cos 5x œ 3h ‰ sin 3h x  x cos x 2 3 lim at
ÚtÛb œ 4
3 œ 1 " ‰ cos 5x xÄ! œ " 3 Œ œ " lim )Ä!c (where ) œ kt) (where ) œ 3y) 3 4 " 3 œ sin ) ) " ‹ Š lim x Ä ! cos 2x xÄ! œ Š #" lim †1œ 2 sin 2x #x ‹ t Ä! " 3 (where ) œ 3h) œ1†2œ2 t " ‹ Š lim cos 5x ‹ x Ä ! sin 2x xÄ! 6x# cos x œ lim ˆ sin xxcos x  œk†1œk tÄ! x Ä ! sin x sin 2x x Ä ! sin x cos x t Ä %c œ œ 2 Š lim cos t‹ Œ lim" sin t  œ 2 † " † " œ 2 t cos t sin t tÄ! xÄ! (b) Ú) Û ) lim 3 sin ) 4 )lim Ä! ) œ œ Š lim sin 2x x Ä ! x cos 2x œ 2 lim xÄ! )Ä! " " sin 3h 3 h lim Ä !c ˆ 3h ‰ œ œ lim œ lim ˆ sinx2x † sin ) ) œ k lim sin 3y 3 4 ylim Ä ! 3y 28. lim 6x# (cot x)(csc 2x) œ lim 29. lim k sin ) ) œ lim 3 sin 3y " 4 ylim 3y Ä! œ h t Ä ! tan t k sin kt kt tÄ! ) Ä $c (where x œ È2)) œ1 sin x x xÄ! œ lim lim h Ä !c sin 3h 26. lim œ lim (b) 2x œ lim ˆ3 cos x † x sin x xÄ! x cos x ‰ sin x cos x œ lim ˆ sinx x † xÄ! † œ ˆ #" † 1‰ (1) œ 2x ‰ sin 2x " ‰ cos x " # œ3†"†1œ3  lim x x Ä ! sin x œ lim Š sin" x ‹ † lim ˆ cos" x ‰  lim Š sin" x ‹ œ (1)(1)  1 œ 2 xÄ! 30. lim xÄ! xÄ! x x#
x  sin x #x 1
cos ) ) Ä ! sin 2) 31. lim œ lim )Ä! )Ä! 34. lim sin (sin h) sin h sin ) ) Ä ! sin 2) 36. lim sin 5x x Ä ! sin 4x œ œ lim )Ä! œ lim )Ä! sin ) ) sin ) ) )Ä!  "# (1) œ 0 1
cos2 ) a2sin ) cos )ba1  cos )b œ lim )Ä! sin2 ) a2sin ) cos )ba1  cos )b œ lim xÄ! xa1 c cos xb 9x2 sin2 3x 9x2 œ lim 1 c cos x 9x 2 x Ä ! ˆ sin3x3x ‰ œ " lim ˆ 1 9 x Ä! cos x ‰ x 2 lim ˆ sin3x3x ‰ xÄ! œ " 9 a0 b 12 œ0 œ 1 since ) œ sin h Ä 0 as h Ä 0 2) ‰ #) 5x œ lim ˆ sin sin 4x † 4x 5x )Ä! œ lim " # œ 1 since ) œ 1
cos t Ä 0 as t Ä 0 sin ) œ lim ˆ sin 2) † xÄ!  "# ˆ sinx x ‰‰ œ 0
œ0 0 a2ba2b xa1
cos xb sin2 3x xÄ! sin(1
cos t) 1
cos t x a1
cos )ba1  cos )b a2sin ) cos )ba1  cos )b œ lim 33. lim 35. lim " # xÄ! œ lim x
x cos x sin2 3x xÄ! hÄ! œ lim ˆ #x
sin ) a2cos )ba1  cos )b 32. lim tÄ! xÄ! œ " # )lim Ä! † 54 ‰ œ ˆ sin) ) † 5 4 xlim Ä! 2) ‰ sin 2) ˆ sin5x5x † œ " # 4x ‰ sin 4x †1†1œ œ 5 4 " # †1†1œ 5 4 Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 65 66 Chapter 2 Limits and Continuity 37. lim ) cos ) œ 0 † 1 œ 0 )Ä! 38. lim sin ) cot 2) œ lim sin ) )Ä! )Ä! tan 3x 39. lim x Ä ! sin 8x œ 3 8 xlim Ä! 40. lim yÄ! œ sin 3x œ lim ˆ cos 3x † xÄ! " ‰ sin 8x œ lim sin ) )Ä! œ lim xÄ! 3 8 †1†1†1œ sin 3y sin 4y cos 5y y cos 4y sin 5y yÄ! œ lim ) cot 4) 2 2 ) Ä ! sin ) cot 2) 42. lim )Ä! sin ) cos ) 3) 2 ) cos sin 3) œ lim )Ä! 2 lim 4) cos2 4) cos ) ) Ä ! cos 2) sin 4) œ lim cos 4) sin 4) 2 2) 2 sin ) cos sin2 2) " sin 8x cos 2) ) Ä ! 2 cos ) † 8x 3x œ 1 2 † 83 ‰ 3 8 yÄ! sin ) sin 3) 2 ) Ä ! ) cos ) cos 3) ) œ lim sin 4y cos 5y 3†4†5y œ lim Š siny3y ‹ Š cos 4y ‹ Š sin 5y ‹ Š 3†4†5y ‹ cos 5y ˆ 3†4 ‰ lim Š sin3y3y ‹ Š sin4y4y ‹ Š sin5y5y ‹ Š cos 4y ‹ 5 yÄ! tan ) 2 ) Ä ! ) cot 3) cos 2) 2sin ) cos ) sin 3x œ lim ˆ cos 3x † ˆ cos"3x ‰ ˆ sin3x3x ‰ ˆ sin8x8x ‰ œ sin 3y cot 5y y cot 4y 41. lim œ cos 2) sin 2) œ1†1†1†1† œ 12 5 œ lim ˆ sin) ) ‰ˆ sin3)3) ‰ˆ cos ) 3cos 3) ‰ œ a1ba1bˆ 13†1 ‰ œ 3 )Ä! ) cos 4) sin2 2) 2 2 ) Ä ! sin ) cos 2) sin 4) œ lim 12 5 ) cos 4) a2sin ) cos )b2 2 2 ) Ä ! sin ) cos 2) sin 4) œ lim ) cos 4) ˆ4sin2 ) cos2 )‰ 2 2 ) Ä ! sin ) cos 2) sin 4) œ lim 4) cos ) 4) cos ) œ lim ˆ sin4)4) ‰Š coscos ‹ œ lim Š sin14) ‹Š coscos ‹ œ ˆ 11 ‰Š 11†12 ‹ œ 1 2 2) 2 2) 2 )Ä! 2 )Ä! 2 4) 43. Yes. If lim b f(x) œ L œ lim c f(x), then xlim f(x) œ L. If lim b f(x) Á lim c f(x), then xlim f(x) does not exist. Äa Äa xÄa xÄa xÄa xÄa 44. Since xlim f(x) œ L if and only if lim b f(x) œ L and lim c f(x) œ L, then xlim f(x) can be found by calculating Äc Äc xÄc xÄc lim b f(x). xÄc 45. If f is an odd function of x, then f(
x) œ
f(x). Given lim b f(x) œ 3, then lim c f(x) œ
$. xÄ! xÄ! 46. If f is an even function of x, then f(
x) œ f(x). Given lim c f(x) œ 7 then xÄ# can be said about lim x Ä
#c lim x Ä
#b f(x) œ 7. However, nothing f(x) because we don't know lim b f(x). xÄ# 47. I œ (5ß 5  $ ) Ê 5  x  &  $ . Also, Èx
5  % Ê x
5  %# Ê x  &  %# . Choose $ œ %# Ê lim Èx
5 œ 0. x Ä &b 48. I œ (%
$ ß %) Ê %
$  x  4. Also, È%
x  % Ê %
x  %# Ê x  %
%# . Choose $ œ %# Ê lim È%
x œ 0. x Ä %c 49. As x Ä 0
the number x is always negative. Thus, ¹ kxxk
(
1)¹  % Ê ¸
xx  1¸  % Ê 0  % which is always true independent of the value of x. Hence we can choose any $  0 with
$  x  ! Ê 2 ¸ x
2 ¸ 50. Since x Ä # we have x  2 and kx
2k œ x
2. Then, ¹ kxx

2 k
" ¹ œ x
2
"  % Ê 0  % which is always true so long as x  #. Hence we can choose any $  !, and thus #  x  #  $ 2 Ê ¹ kxx

2k
"¹  % . Thus, x
2 lim x Ä
#b kx
2k x lim x Ä ! c kx k œ 1. Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. œ
1. Section 2.5 Continuity 51. (a) (b) lim x Ä %!!b 67 ÚxÛ œ 400. Just observe that if 400  x  401, then ÚxÛ œ 400. Thus if we choose $ œ ", we have for any number %  ! that 400  x  400  $ Ê lÚxÛ
400l œ l400
400l œ !  %. lim c ÚxÛ œ 399. Just observe that if 399  x  400 then ÚxÛ œ 399. Thus if we choose $ œ ", we have for any x Ä %!! number %  ! that 400
$  x  400 Ê lÚxÛ
399l œ l399
399l œ !  %. (c) Since lim b ÚxÛ Á lim c ÚxÛ we conclude that lim ÚxÛ does not exist. x Ä %!! x Ä %!! 52. (a) x Ä %!! lim f(x) œ lim b Èx œ È0 œ 0; ¸Èx
0¸  % Ê
%  Èx  % Ê !  x  %# for x positive. Choose $ œ %# xÄ! Ê lim b f(x) œ 0. x Ä !b xÄ! (b) lim f(x) œ lim c x# sin ˆ x" ‰ œ 0 by the sandwich theorem since
x# Ÿ x# sin ˆ x" ‰ Ÿ x# for all x Á 0. x Ä !c xÄ! Since kx#
0k œ k
x#
0k œ x#  % whenever kxk  È%, we choose $ œ È% and obtain ¸x# sin ˆ "x ‰
0¸  % if
$  x  0. (c) The function f has limit 0 at x! œ 0 since both the right-hand and left-hand limits exist and equal 0. 2.5 CONTINUITY 1. No, discontinuous at x œ 2, not defined at x œ 2 2. No, discontinuous at x œ 3, " œ lim c g(x) Á g(3) œ 1.5 xÄ$ 3. Continuous on [
1ß 3] 4. No, discontinuous at x œ 1, 1.5 œ lim c k(x) Á lim b k(x) œ ! xÄ" xÄ" 5. (a) Yes (b) Yes, (c) Yes (d) Yes 6. (a) Yes, f(1) œ 1 lim x Ä
"b f(x) œ 0 (b) Yes, lim f(x) œ 2 xÄ1 (c) No (d) No 7. (a) No (b) No 8. [
"ß !)  (!ß ")  ("ß #)  (#ß $) 9. f(2) œ 0, since lim c f(x) œ
2(2)  4 œ 0 œ lim b f(x) xÄ# xÄ# 10. f(1) should be changed to 2 œ lim f(x) xÄ1 11. Nonremovable discontinuity at x œ 1 because lim f(x) fails to exist ( limc f(x) œ 1 and lim b f(x) œ 0). xÄ" xÄ1 xÄ" Removable discontinuity at x œ 0 by assigning the number lim f(x) œ 0 to be the value of f(0) rather than f(0) œ 1. xÄ! 12. Nonremovable discontinuity at x œ 1 because lim f(x) fails to exist ( lim c f(x) œ 2 and lim b f(x) œ 1). xÄ" xÄ1 xÄ" Removable discontinuity at x œ 2 by assigning the number lim f(x) œ 1 to be the value of f(2) rather than f(2) œ 2. xÄ# Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 68 Chapter 2 Limits and Continuity 13. Discontinuous only when x
2 œ 0 Ê x œ 2 14. Discontinuous only when (x  2)# œ 0 Ê x œ
2 15. Discontinuous only when x#
%x  $ œ ! Ê (x
3)(x
1) œ 0 Ê x œ 3 or x œ 1 16. Discontinuous only when x#
3x
10 œ 0 Ê (x
5)(x  2) œ 0 Ê x œ 5 or x œ
2 17. Continuous everywhere. ( kx
1k  sin x defined for all x; limits exist and are equal to function values.) 18. Continuous everywhere. ( kxk  " Á 0 for all x; limits exist and are equal to function values.) 19. Discontinuous only at x œ 0 20. Discontinuous at odd integer multiples of 1# , i.e., x = (2n
") 1# , n an integer, but continuous at all other x. 21. Discontinuous when 2x is an integer multiple of 1, i.e., 2x œ n1, n an integer Ê x œ n1 # , n an integer, but continuous at all other x. 22. Discontinuous when 1x # is an odd integer multiple of 1# , i.e., 1x # œ (2n
1) 1# , n an integer Ê x œ 2n
1, n an integer (i.e., x is an odd integer). Continuous everywhere else. 23. Discontinuous at odd integer multiples of 1# , i.e., x = (2n
1) 1# , n an integer, but continuous at all other x. 24. Continuous everywhere since x%  1 and are equal to the function values. 1 and
" Ÿ sin x Ÿ 1 Ê 0 Ÿ sin# x Ÿ 1 Ê 1  sin# x 1; limits exist 25. Discontinuous when 2x  3  0 or x 
3# Ê continuous on the interval

3# ß _‰ . 26. Discontinuous when 3x
1  0 or x  " 3 Ê continuous on the interval
3" ß _‰ . 27. Continuous everywhere: (2x
1)"Î$ is defined for all x; limits exist and are equal to function values. 28. Continuous everywhere: (2
x)"Î& is defined for all x; limits exist and are equal to function values. 29. Continuous everywhere since lim xÄ3 30. Discontinuous at x œ
2 since x2
x
6 x
3 œ lim xÄ3 ax
3bax  2b x
3 œ lim ax  2b œ 5 œ ga3b xÄ3 lim faxb does not exist while fa
2b œ 4. x Ä
2 31. xlim sin (x
sin x) œ sin (1
sin 1) œ sin (1
0) œ sin 1 œ 0, and function continuous at x œ 1. Ä1 32. lim sin ˆ 1# cos (tan t)‰ œ sin ˆ 1# cos (tan (0))‰ œ sin ˆ 1# cos (0)‰ œ sin ˆ 1# ‰ œ 1, and function continuous at t œ !. tÄ! 33. lim sec ay sec# y
tan# y
1b œ lim sec ay sec# y
sec# yb œ lim sec a(y
1) sec# yb œ sec a("
") sec# 1b yÄ1 yÄ1 yÄ1 œ sec 0 œ 1, and function continuous at y œ ". 34. lim tan
14 cos ˆsin x"Î$ ‰‘ œ tan
14 cos (sin(0))‘ œ tan ˆ 14 cos (0)‰ œ tan ˆ 14 ‰ œ 1, and function continuous at x œ !. xÄ! Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Section 2.5 Continuity 35. lim cos ’ È19
13 sec 2t “ œ cos ’ È19
13 sec 0 “ œ cos tÄ! 1 È16 œ cos 1 4 œ È2 # , and function continuous at t œ !. 36. lim1 Écsc# x  5È3 tan x œ Écsc# ˆ 16 ‰  5È3 tan ˆ 16 ‰ œ Ê4  5È3 Š È"3 ‹ œ È9 œ 3, and function continuous at xÄ ' x œ 1' . 37. g(x) œ x#
9 x
3 (x  3)(x
3) (x
3) œ 38. h(t) œ t#  3t
10 t
# 39. f(s) œ s$
" s#
1 40. g(x) œ œ œ œ x  3, x Á 3 Ê g(3) œ lim (x  3) œ 6 xÄ$ (t  5)(t
2) t
# as#  s  1b (s
1) (s  1)(s
1) x#
16 x#
3x
4 œ œ t  5, t Á # Ê h(2) œ lim (t  5) œ 7 tÄ# s#  s  " s1 , œ (x  4)(x
4) (x
4)(x  1) œ x4 x1 s Á 1 Ê f(1) œ lim Š s sÄ1 # s1 s1 ‹ 4‰ , x Á 4 Ê g(4) œ lim ˆ xx  1 œ xÄ% œ 3 # 8 5 41. As defined, lim c f(x) œ (3)#
1 œ 8 and lim b (2a)(3) œ 6a. For f(x) to be continuous we must have xÄ$ xÄ$ 6a œ 8 Ê a œ 43 . 42. As defined, lim x Ä
#c g(x) œ
2 and 4b œ
2 Ê b œ
"# . lim x Ä
#b g(x) œ b(
2)# œ 4b. For g(x) to be continuous we must have 43. As defined, lim c f(x) œ 12 and lim b f(x) œ a# a2b
2a œ 2a#
2a. For f(x) to be continuous we must have xÄ# xÄ# 12 œ 2a#
2a Ê a œ 3 or a œ
2. 44. As defined, lim c g(x) œ
b b1 xÄ0 0
b b1 œ
b b1 œ b Ê b œ 0 or b œ
2. 45. As defined, lim x Ä
1 c f(x) œ
2 and and lim b g(x) œ a0b2  b œ b. For g(x) to be continuous we must have xÄ0 lim x Ä
1 b f(x) œ aa
1b  b œ
a  b, and lim f(x) œ aa1b  b œ a  b and x Ä 1c lim f(x) œ 3. For f(x) to be continuous we must have
2 œ
a  b and a  b œ 3 Ê a œ x Ä 1b 5 # and b œ "# . 46. As defined, lim c g(x) œ aa0b  2b œ 2b and lim b g(x) œ a0b2  3a
b œ 3a
b, and xÄ0 xÄ0 lim g(x) œ a2b2  3a
b œ 4  3a
b and lim b g(x) œ 3a2b
5 œ 1. For g(x) to be continuous we must xÄ0 x Ä 2c have 2b œ 3a
b and 4  3a
b œ 1 Ê a œ
3# and b œ
3# . Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 69 70 Chapter 2 Limits and Continuity 47. The function can be extended: f(0) ¸ 2.3. 48. The function cannot be extended to be continuous at x œ 0. If f(0) ¸ 2.3, it will be continuous from the right. Or if f(0) ¸
2.3, it will be continuous from the left. 49. The function cannot be extended to be continuous at x œ 0. If f(0) œ 1, it will be continuous from the right. Or if f(0) œ
1, it will be continuous from the left. 50. The function can be extended: f(0) ¸ 7.39. 51. f(x) is continuous on [!ß "] and f(0)  0, f(1)  0 Ê by the Intermediate Value Theorem f(x) takes on every value between f(0) and f(1) Ê the equation f(x) œ 0 has at least one solution between x œ 0 and x œ 1. 52. cos x œ x Ê (cos x)
x œ 0. If x œ
1# , cos ˆ
1# ‰
ˆ
1# ‰  0. If x œ 1# , cos ˆ 1# ‰
for some x between
1 # and 1 # 1 #  0. Thus cos x
x œ 0 according to the Intermediate Value Theorem, since the function cos x
x is continuous. 53. Let f(x) œ x$
15x  1, which is continuous on [
4ß 4]. Then f(
4) œ
3, f(
1) œ 15, f(1) œ
13, and f(4) œ 5. By the Intermediate Value Theorem, f(x) œ 0 for some x in each of the intervals
%  x 
1,
"  x  1, and "  x  4. That is, x$
15x  1 œ 0 has three solutions in [
%ß 4]. Since a polynomial of degree 3 can have at most 3 solutions, these are the only solutions. 54. Without loss of generality, assume that a  b. Then F(x) œ (x
a)# (x
b)#  x is continuous for all values of x, so it is continuous on the interval [aß b]. Moreover F(a) œ a and F(b) œ b. By the Intermediate Value Theorem, since a  a # b  b, there is a number c between a and b such that F(x) œ a # b . Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Section 2.5 Continuity 55. Answers may vary. Note that f is continuous for every value of x. (a) f(0) œ 10, f(1) œ 1$
8(1)  10 œ 3. Since $  1  10, by the Intermediate Value Theorem, there exists a c so that !  c  1 and f(c) œ 1. (b) f(0) œ 10, f(
4) œ (
4)$
8(
4)  10 œ
22. Since
22 
È3  10, by the Intermediate Value Theorem, there exists a c so that
4  c  0 and f(c) œ
È3. (c) f(0) œ 10, f(1000) œ (1000)$
8(1000)  10 œ 999,992,010. Since 10  5,000,000  999,992,010, by the Intermediate Value Theorem, there exists a c so that !  c  1000 and f(c) œ 5,000,000. 56. All five statements ask for the same information because of the intermediate value property of continuous functions. (a) A root of f(x) œ x$
3x
1 is a point c where f(c) œ 0. (b) The points where y œ x$ crosses y œ 3x  1 have the same y-coordinate, or y œ x$ œ 3x  1 Ê f(x) œ x$
3x
1 œ 0. (c) x$
3x œ 1 Ê x$
3x
1 œ 0. The solutions to the equation are the roots of f(x) œ x$
3x
1. (d) The points where y œ x$
3x crosses y œ 1 have common y-coordinates, or y œ x$
3x œ 1 Ê f(x) œ x$
3x
1 œ !. (e) The solutions of x$
3x
1 œ 0 are those points where f(x) œ x$
3x
1 has value 0. 57. Answers may vary. For example, f(x) œ sin (x
2) x
2 is discontinuous at x œ 2 because it is not defined there. However, the discontinuity can be removed because f has a limit (namely 1) as x Ä 2. 58. Answers may vary. For example, g(x) œ " x1 has a discontinuity at x œ
1 because lim g(x) does not exist. x Ä
" Š lim c g(x) œ
_ and lim b g(x) œ _.‹ x Ä
" x Ä
" 59. (a) Suppose x! is rational Ê f(x! ) œ 1. Choose % œ "# . For any $  0 there is an irrational number x (actually infinitely many) in the interval (x!
$ ß x!  $ ) Ê f(x) œ 0. Then 0  kx
x! k  $ but kf(x)
f(x! )k œ 1  "# œ %, so x lim f(x) fails to exist Ê f is discontinuous at x! rational. Äx ! On the other hand, x! irrational Ê f(x! ) œ 0 and there is a rational number x in (x!
$ ß x!  $ ) Ê f(x) œ 1. Again x lim f(x) fails to exist Ê f is discontinuous at x! irrational. That is, f is discontinuous at Äx ! every point. (b) f is neither right-continuous nor left-continuous at any point x! because in every interval (x!
$ ß x! ) or (x! ß x!  $ ) there exist both rational and irrational real numbers. Thus neither limits lim f(x) and x Ä x! lim f(x) exist by the same arguments used in part (a). x Ä x
! 60. Yes. Both f(x) œ x and g(x) œ x
g ˆ "# ‰ œ0 Ê f(x) g(x) " # are continuous on [!ß "]. However is discontinuous at x œ f(x) g(x) is undefined at x œ " # since " #. 61. No. For instance, if f(x) œ 0, g(x) œ ÜxÝ, then h(x) œ 0 aÜxÝb œ 0 is continuous at x œ 0 and g(x) is not. 62. Let f(x) œ œ " x
1 " (x  1)
1 œ and g(x) œ x  1. Both functions are continuous at x œ 0. The composition f ‰ g œ f(g(x)) " x is discontinuous at x œ 0, since it is not defined there. Theorem 10 requires that f(x) be continuous at g(0), which is not the case here since g(0) œ 1 and f is undefined at 1. 63. Yes, because of the Intermediate Value Theorem. If f(a) and f(b) did have different signs then f would have to equal zero at some point between a and b since f is continuous on [aß b]. Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 71 72 Chapter 2 Limits and Continuity 64. Let f(x) be the new position of point x and let d(x) œ f(x)
x. The displacement function d is negative if x is the left-hand point of the rubber band and positive if x is the right-hand point of the rubber band. By the Intermediate Value Theorem, d(x) œ 0 for some point in between. That is, f(x) œ x for some point x, which is then in its original position. 65. If f(0) œ 0 or f(1) œ 1, we are done (i.e., c œ 0 or c œ 1 in those cases). Then let f(0) œ a  0 and f(1) œ b  1 because 0 Ÿ f(x) Ÿ 1. Define g(x) œ f(x)
x Ê g is continuous on [0ß 1]. Moreover, g(0) œ f(0)
0 œ a  0 and g(1) œ f(1)
1 œ b
1  0 Ê by the Intermediate Value Theorem there is a number c in (!ß ") such that g(c) œ 0 Ê f(c)
c œ 0 or f(c) œ c. 66. Let % œ kf(c)k #  0. Since f is continuous at x œ c there is a $  0 such that kx
ck  $ Ê kf(x)
f(c)k  % Ê f(c)
%  f(x)  f(c)  %. If f(c)  0, then % œ "# f(c) Ê " # " # If f(c)  0, then % œ
f(c) Ê f(c)  f(x)  3 # 3 # f(c)  f(x)  f(c) Ê f(x)  0 on the interval (c
$ ß c  $ ). " # f(c) Ê f(x)  0 on the interval (c
$ ß c  $ ). 67. By Exercises 52 in Section 2.3, we have xlim faxb œ L Í lim fac  hb œ L. Äc hÄ0 Thus, faxb is continuous at x œ c Í xlim faxb œ facb Í lim fac  hb œ facb. Äc hÄ0 68. By Exercise 67, it suffices to show that lim sinac  hb œ sin c and lim cosac  hb œ cos c. hÄ0 hÄ0 Now lim sinac  hb œ lim
asin cbacos hb  acos cbasin hb‘ œ asin cbŠ lim cos h‹  acos cbŠ lim sin h‹ hÄ0 hÄ0 hÄ0 hÄ0 By Example 11 Section 2.2, lim cos h œ " and lim sin h œ !. So lim sinac  hb œ sin c and thus faxb œ sin x is hÄ0 continuous at x œ c. Similarly, hÄ0 hÄ0 lim cosac  hb œ lim
acos cbacos hb
asin cbasin hb‘ œ acos cbŠ lim cos h‹
asin cbŠ lim sin h‹ œ cos c. hÄ0 hÄ0 Thus, gaxb œ cos x is continuous at x œ c. hÄ0 69. x ¸ 1.8794,
1.5321,
0.3473 70. x ¸ 1.4516,
0.8547, 0.4030 71. x ¸ 1.7549 72. x ¸ 1.5596 73. x ¸ 3.5156 74. x ¸
3.9058, 3.8392, 0.0667 75. x ¸ 0.7391 76. x ¸
1.8955, 0, 1.8955 hÄ0 Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Section 2.6 Limits Involving Infinity; Asymptotes of Graphs 2.6 LIMITS INVOLVING INFINITY; ASMYPTOTES OF GRAPHS 1. (a) (c) (e) (g) lim f(x) œ 0 (b) xÄ2 lim x Ä
3 c f(x) œ 2 lim x Ä
3 b f(x) œ
2 (d) lim f(x) œ does not exist xÄ3 lim b f(x) œ
1 (f) lim f(x) œ does not exist (h) x lim f(x) œ 1 Ä_ xÄ0 xÄ0 lim f(x) œ  _ x Ä 0c (i) x Ä lim f(x) œ 0
_ 2. (a) (c) lim f(x) œ 2 (b) lim f(x) œ 1 (d) lim f(x) œ does not exist xÄ4 x Ä 2c xÄ2 lim f(x) œ  _ x Ä
3 b (g) lim f(x) œ  _ (e) x Ä
3 (i) lim f(x) œ
3 x Ä 2b lim f(x) œ
_ (f) x Ä
3 c lim (h) x Ä 0b (k) x lim f(x) œ 0 Ä_ lim f(x) œ  _ lim f(x) œ does not exist (j) x Ä 0c xÄ0 (l) x Ä lim f(x) œ
1
_ Note: In these exercises we use the result " lim mÎn xÄ „_ x Theorem 8 and the power rule in Theorem 1: lim xÄ „_ œ 0 whenever ˆ xm"În ‰ œ lim (b)
3 4. (a) 1 (b) 1 5. (a) " # (b) " # 6. (a) " 8 (b) " 8 7. (a)
53 (b) 10.
3") Ÿ 11. lim tÄ_ 12. r Ä lim_  0. This result follows immediately from ˆ x" ‰mÎn œ Š " lim ‹ xÄ „_ x mÎn (b)
53 3 4 9.
"x Ÿ m n xÄ „_ 3. (a)
3 8. (a) f(x) œ  _ sin 2x x Ÿ " x cos ) 3) Ÿ " 3) 2
t  sin t t  cos t Ê x lim Ä_ Ê 13. (a) x lim Ä_ 2x  3 5x  7 lim ) Ä
_ œ lim 2 t tÄ_ r  sin r 2r  7
5 sin r sin 2x x œrÄ lim_ œ x lim Ä_ œ 0 by the Sandwich Theorem cos ) 3) œ 0 by the Sandwich Theorem
1  ˆ sint t ‰ 1  ˆ cost t ‰ œ 1  ˆ sinr r ‰ 2  7r
5 ˆ sinr r ‰ 2  3x 5  7x 3 4 œ 2 5 0
10 10 œ
1 œrÄ lim_ 10 20
0 œ (b) " # 2 5 (same process as part (a)) Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. œ 0mÎn œ 0. 73 74 Chapter 2 Limits and Continuity 2  Š x7$ ‹ $ 2x  7 14. (a) x lim œ x lim Ä _ x$
x#  x  7 Ä_ (b) 2 (same process as part (a)) " x  x"# 15. (a) x lim Ä_ x1 x#  3 œ x lim Ä_ 1  x3# 16. (a) x lim Ä_ 3x  7 x#
2 œ x lim Ä_ 1
x2# 17. (a) x lim Ä_ 7x$ x$
3x#  6x 18. (a) x lim Ä_ x$ 20. (a) x lim Ä_ 9 # 2x%  x7# œ x lim Ä_ œ x lim Ä_ 10x&  x%  31 x' 19. (a) x lim Ä_ (b) "
4x  1 3 x œ2 œ0 (b) 0 (same process as part (a)) œ0 (b) 0 (same process as part (a)) œ( 7 1
3x  x6# " x$ 4  x"$ x# 1
œ x lim Ä_ 2 (b) 7 (same process as part (a)) œ!  x"#  x31' 1 10 x œ x lim Ä_ 9x%  x  5x#
x  6 1
"x  x"#  x7$ (b) 0 (same process as part (a)) œ0 (b) 0 (same process as part (a)) 9  x"$ 5 x#
x"$  x6% œ 9 # (same process as part (a))
2x$
2x  3 3x$  3x#
5x 21. (a) x lim Ä_
2
x2#  x3$ œ x lim Ä_ œ
23 3  3x
x5# (b)
23 (same process as part (a))
x % x%
7x$  7x#  9 22. (a) x lim Ä_
" 1
7x  x7#  x9% œ x lim Ä_ œ
1 (b)
1 (same process as part (a)) 8
8
3 3
3 x2 x2 É 8x 23. x lim œ Êx lim œ É 82
00 œ È4 œ 2 2x2  x œ x lim Ä_ Ä _ Ê 2  1x Ä _ 2  1x 2 24. x Ä lim Šx x
1‹
_ 8x2
3 2 1 Î3 œxÄ lim
_Π5 1 "  1x
x12 8

x 3 x2 1 Î3  œ Œx Ä lim
_ 5 1 "  1x
x12 8
3 x2 5
x 1 Î3  œ ˆ " 8
0
0 0 ‰ 1 Î3 œ ˆ "8 ‰ 1 Î3 œ x2 x2 ‰ œ_ 25. x Ä lim lim œ Œx Ä lim œ ˆ 01
_ Š 1
x ‹ œ x Ä 0
_ x2
7x
_Π1
7x 
_ 1
7x  3 1
5 1
5 5 x x x2 x2 É x
5x œ x lim 26. x lim œ Êx lim œ É 1 0
0
0 0 œ È0 œ 0 Ä _ x3  x
2 Ä _Ê 1  x12
x23 Ä _ 1  x12
x23 2 27. x lim Ä_ 2Èx  xc" 3x
7 29. x Ä lim
_ 30. x lim Ä_ œ x lim Ä_ 3 x
È 5 x È 3 xÈ 5 x È x "x % x #
x $ Š œxÄ lim
_ œ x lim Ä_ 2 ‹  Š x"# ‹ x"Î# 3
7x œ0 1
xÐ"Î&Ñ Ð"Î$Ñ 1  xÐ"Î&Ñ Ð"Î$Ñ x  x"# 1
x" œxÄ lim
_ 28. x lim Ä_ " ‹ 1
Š #Î"& x " ‹ 1  Š #Î"& 2  Èx 2
Èx œ x lim Ä_ 2 ‹" x"Î# 2 Š "Î# ‹
1 x Š œ
1 œ1 x œ_ Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. " # Section 2.6 Limits Involving Infinity; Asymptotes of Graphs 2x&Î$
x"Î$  7 x)Î&  3x  Èx 31. x lim Ä_ 3 x
5x  3 È 2x  x#Î$
4 32. x Ä lim
_ 33. x lim Ä_ È x2  1 x1 34. x Ä lim
_ œ x lim Ä_ œxÄ lim
_ œ x lim Ä_ È x2  1 x1 1 " x#Î$ 2 È x 2  1 ÎÈ x 2 a x  1 b ÎÈ x 2 œxÄ lim
_ "  7 x"*Î"& x)Î& 3 "  x$Î& x""Î"! 2x"Î"&

5  3x " x"Î$ È 3 œ
5# È x
4x œ x lim Ä_ È x 2  1 ÎÈ x 2 a x  1 bÎ È x 2 œ_ È a x 2  1 bÎx 2 ax  1bÎx œ x lim Ä_ È 1  1 Îx 2 a1  1 Îx b È a x 2  1 bÎ x 2 È1  0 a1  0 b œ È 1  1 Îx 2 œxÄ lim œ x lim œ
_ ax  1bÎa
xb Ä _ a
1
1 Î x b œ1 È1  0 a
1
0b a x
3 bÎ x a x
3 bÎ x a1
3 Îx b x
3 35. x lim œ x lim œ x lim œ x lim œ Ä _ È4x2  25 Ä _ È4x2  25ÎÈx2 Ä _ Èa4x2  25bÎx2 Ä _ È4  25Îx2 ˆ4
3x3 ‰ÎÈx6 4
3x 36. x Ä lim œxÄ lim œxÄ lim
_ Èx6  9
_ Èx6  9ÎÈx6
_ 3 " œ_ 37. lim x Ä !b 3x 39. lim x Ä #c x
2 41. lim x Ä
)b x8 3 2x 4 43. lim # x Ä ( (x
7) œ
_ œ
_ œ_ lim "Î$ x Ä !b 3x 46. (a) lim "Î& x Ä !b x 4 47. lim #Î& xÄ! x 49. 51. 52. lim x Ä ˆ 1# ‰ œ lim 4 # x Ä ! ax"Î& b œ x lim Ä_ ˆ
4 Îx 3  3‰ È 1  9 Îx 6 Š positive positive ‹ 40. lim x Ä $b x
3 Š negative positive ‹ 42. lim x Ä
&c 2x10 positive Š positive ‹ 44. lim œ_ " 3x
" # x Ä ! x (x1) 2 (b) lim "Î$ x Ä !c 3x (b) lim "Î& x Ä !c x 48. lim " #Î$ xÄ! x tan x œ _ 50. œ3 œ_ positive Š negative ‹ 2 a0  3 b È1  0 " # positive Š negative ‹ lim x Ä !c 2x 5 œ œ œ
_ 38. œ_ 2 ˆ4
3x3 ‰Îˆ
x3 ‰ Èax6  9bÎx6 positive Š positive ‹ œ_ 2 45. (a) a1
0 b È4  0 2 œ
1 œ_ Š negative negative ‹ œ
_ negative Š positive †positive ‹ œ
_ œ
_ œ lim " # x Ä ! ax"Î$ b œ_ lim
sec x œ _ x Ä ˆ #1 ‰ lim (1  csc )) œ
_ )Ä! lim (2
cot )) œ
_ and lim c (2
cot )) œ _, so the limit does not exist )Ä! ) Ä !b " œ lim b xÄ# " (x2)(x
2) œ_ Š positive"†positive ‹ " œ lim c xÄ# " (x2)(x
2) œ
_ Š positive†"negative ‹ 53. (a) lim # x Ä # b x
4 (b) lim # x Ä # c x
4 (c) lim # x Ä
#b x
4 (d) lim # x Ä
#c x
4 " œ lim x Ä
#b (x2)(x
2) " œ
_ Š positive†"negative ‹ " œ lim x Ä
#c (x2)(x
2) " œ_ Š negative"†negative ‹ Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 75 76 Chapter 2 Limits and Continuity 54. (a) lim # x Ä "b x
1 (b) lim # x Ä "c x
1 (c) lim # x Ä
"b x
1 (d) lim # x Ä
"c x
1 x œ lim b xÄ" x (x1)(x
1) œ_ positive Š positive †positive ‹ x œ lim c xÄ" x (x1)(x
1) œ
_ positive Š positive †negative ‹ x œ lim x Ä
"b (x1)(x
1) x œ_ negative Š positive †negative ‹ x œ lim x Ä
"c (x1)(x
1) x œ
_ negative Š negative †negative ‹ 55. (a) lim x Ä !b # x#
" x œ 0  lim b xÄ! "
x œ
_ " Š negative ‹ (b) lim x Ä !c # x#
" x œ 0  lim c xÄ! "
x œ_ " Š positive ‹ (c) lim # x Ä $È2 (d) lim x Ä
1 # 56. (a) x# x# lim x Ä
#b " x

" x œ x#
1 2x  4 x#
1 2#Î$ # œ " # (d) lim x Ä !c 2x  4 œ (b) (c) (d) (e) 58. (a) x#
3x  2 x$
2x# lim b x#
3x  2 x$
2x# lim xÄ# # x
3x  2 x$
2x# x
3x  2 x$
2x# lim œ lim c xÄ# lim x Ä
#b (c) x Ä 0c (d) x Ä "b (e) lim x Ä !b x(x  #) x#
1 2x  4 lim x Ä
#c œ
_ positive Š negative ‹ œ0 (x
2)(x
1) x# (x
2) œ
_ (x
2)(x
1) x# (x
2) œ lim b xÄ# (x
2)(x
1) x# (x
2) œ lim c xÄ# œ lim œ lim (x
2)(x
1) x# (x
2) œ
_ xÄ! lim x#
3x  2 x$
4x lim x#
3x  2 x$
4x x
" 2†0 #4 (x
2)(x
1) x# (x
2) xÄ# œ lim b xÄ# x#
3x  2 x$
4x (b) œ œ lim x#
3x  2 x$
4x lim x Ä #b and œ lim b xÄ# x
3x  2 x$
2x# # xÄ! œ lim b xÄ! # lim x Ä #c (x  1)(x
1) 2x  4 (b)
" 4 lim b xÄ# 3 # Š positive positive ‹ œ lim b xÄ" xÄ! œ 2
"Î$
2
"Î$ œ 0 œ_ lim x Ä "b 2x  4 57. (a) " #"Î$
ˆ
"1 ‰ œ (c) x#
1
œ xÄ# (x
2)(x
") x(x
#)(x  2) (x
2)(x
") œ lim c xÄ! œ lim b xÄ" (x
2)(x
") x(x
#)(x  2) (x
2)(x
") x(x
#)(x  2) œ x
1 x# œ " 4 ,xÁ2 x
1 x# œ " 4 ,xÁ2 x
1 x# œ " 4 ,xÁ2 †negative Š negative positive†negative ‹ œ lim b xÄ# lim x Ä
#b x(x
#)(x  2) †negative Š negative positive†negative ‹ (x
1) x(x  #) œ (x
1) lim x Ä
#b x(x  #) œ lim c xÄ! œ lim b xÄ" œ_ (x
1) x(x  #) œ negative Š positive †positive ‹ x
" negative Š negative †positive ‹ œ_ œ " 8 œ_ (x
1) x(x  #) œ
_ lim x Ä !c x(x  #) " #(4) 0 (1)(3) negative Š negative †positive ‹ negative Š negative †positive ‹ œ0 so the function has no limit as x Ä 0. lim
2
59. (a) t Ä !b 60. (a) t Ä !b 61. (a) x Ä !b (c) x Ä "b 3 ‘ t"Î$ œ
_ " lim
t$Î&  7‘ œ _ lim
2
lim
lim "  ’ x#Î$ 2 “ (x
1)#Î$ œ_ lim "  ’ x#Î$ 2 “ (x
1)#Î$ œ_ (b) t Ä !c (b) t Ä !c lim "  ’ x#Î$ 2 “ (x
1)#Î$ œ_ (b) x Ä !c lim "  ’ x#Î$ 2 “ (x
1)#Î$ œ_ (d) x Ä "c " t$Î& 3 ‘ t"Î$ œ_  7‘ œ
_ Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. Section 2.6 Limits Involving Infinity; Asymptotes of Graphs lim "
’ x"Î$ 1 “ (x
1)%Î$ œ_ (b) x Ä !c lim "
’ x"Î$ 1 “ (x
1)%Î$ œ
_ (d) x Ä "c 62. (a) x Ä !b (c) x Ä "b lim "
’ x"Î$ 1 “ (x
1)%Î$ œ
_ lim "
’ x"Î$ 1 “ (x
1)%Î$ œ
_ 63. y œ " x
1 64. y œ " x1 65. y œ " #x  4 66. y œ
3 x
3 67. y œ x3 x2 68. y œ 2x x1 œ1 " x# 69. Here is one possibility. œ#
2 x1 70. Here is one possibility. Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 77 78 Chapter 2 Limits and Continuity 71. Here is one possibility. 72. Here is one possibility. 73. Here is one possibility. 74. Here is one possibility. 75. Here is one possibility. 76. Here is one possibility. 77. Yes. If x lim Ä_ f(x) g(x) œ 2 then the ratio of the polynomials' leading coefficients is 2, so x Ä lim
_ f(x) g(x) œ 2 as well. 78. Yes, it can have a horizontal or oblique asymptote. 79. At most 1 horizontal asymptote: If x lim Ä_ f(x) lim x Ä
_ g(x) f(x) g(x) œ L, then the ratio of the polynomials' leading coefficients is L, so œ L as well. Èx  9  Èx  4 80. x lim Š Èx  9
Èx  4‹ œ x lim ’Èx  9
Èx  4“ † ’ Èx  9  Èx  4 “ œ x lim Ä_ Ä_ Ä_ 5 Èx 5 0 œ x lim œ x lim œ 11 œ 0 9 4 Ä _ Èx  9  Èx  4 Ä_ ax  9 b
a x  4 b Èx  9  Èx  4 É1  x  É1  x È 2 È 2 81. x lim Š Èx2  25
Èx2
"‹ œ x lim ’Èx2  25
Èx2
"“ † ’ Èx2  25  Èx2
" “ œ x lim Ä_ Ä_ Ä_ x  25  x
" œ x lim Ä_ 26 Èx2  25  Èx2
" œ x lim Ä_ 26 x É1  x252  É1
x12 œ 0 11 œ0 È 2 ˆx  3 ‰
ˆ x ‰
x 82. x Ä lim lim lim Š Èx2  3  x‹ œ x Ä ’Èx2  3  x“ † ’ Èxx2  33
“œxÄ
_
_
_ Èx2  3
x x 3 È x2
3x 3 œxÄ lim œ lim œ lim œ 1 0 1 œ 0 2 È 3 x
_ x Ä
_ É1  2
È x Ä
_ É1  32  1 x 3
x x x x2 2 2 Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley. ˆx2  25‰
ˆx2
"‰ Èx2  25  Èx2
" Section 2.6 Limits Involving Infinity; Asymptotes of Graphs È ˆ4x2 ‰
ˆ4x2  3x
2‰
4x  3x
2 83. x Ä lim lim lim Š 2x  È4x2  3x
2‹ œ x Ä ’2x  È4x2  3x
2“ † ’ 2x “œxÄ
_
_
_ 2x
È4x2  3x
2 2x
È4x2  3x
2 c3x b 2 c3x b 2 3
x2 Èx2
3x  2 cx œxÄ lim œxÄ lim œxÄ lim œxÄ lim
_ 2x
È4x2  3x
2
_ È2x
É4  3
22
_ 2x
É4  3
22
_
2
É4  3
22 x x x cx x x x x2 œ
32

02 œ
43 2 È 2 84. x lim Š È9x2
x
3x‹ œ œ x lim ’È9x2
x
3x“ † ’ È9x2
x  3x “ œ x lim Ä_ Ä_ Ä_ 9x
x  3x œ x lim Ä_
x È9x2
x  3x œ x lim Ä_
xx 2 É 9x2 x
xx2  3x x
1 É9
"x  3 œ x lim Ä_ œ
1 33 ˆ9x2
x‰
ˆ9x2 ‰ È9x2
x  3x œ
"6 È 2 È 2 85. x lim Š Èx2  3x
Èx2
2x‹ œ x lim ’ Èx2  3x
Èx2
2x“ † ’ Èx2  3x  Èx2
2x “ œ x lim Ä_ Ä_ Ä_ x  3x  x
2x 5x 5 5 5 œ x lim œ x lim œ 11 œ # È 2 3 2 Ä_ È 2 Ä_ x  3x  x
2x É1  x  É1
x Èx#  x
Èx#
x œ lim ’Èx#  x
Èx#
x“ † ’ Èx#  x  Èx#
x “ œ lim 86. x lim È x#  x  È x#
x Ä_ xÄ_ xÄ_ 2x 2 2 œ x lim œ x lim œ 11 œ 1 È # " " Ä_ È # Ä_ x x ˆx2  3x‰
ˆx2
2x‰ Èx2  3x  Èx2
2x x
x ax #  x b
a x #
x b È x#  x  È x#
x É1  x  É1
x 87. For any %  0, take N œ 1. Then for all x  N we have that kf(x)
kk œ kk
kk œ 0  %. 88. For any %  0, take N œ 1. Then for all y 
N we have that kf(x)
kk œ kk
kk œ 0  %.
" x# 89. For every real number
B  0, we must find a $  0 such that for all x, 0  kx
0k  $ Ê
" x# Ê 
B