Single variable calculus concepts and contexts 4th edition stewart solutions manual

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Single Variable Calculus Concepts and Contexts 4th Edition Stewart Solutions Manual Visit to Download in Full: https://testbankdeal.com/download/single-variable-calculusconcepts-and-contexts-4th-edition-stewart-solutions-manual/
COMPLETE SOLUTIONS MANUAL for Stewart’s SINGLE VARIABLE CALCULUS CONCEPTS AND CONTEXTS FOURTH EDITION Australia Brazil Japan Korea Mexico Singapore Spain UnitedKingdom United States JEFFERY A. COLE Anoka Ramsey Community College TIMOTHY J. FLAHERTY Carnegie Mellon University NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved Single Variable Calculus Concepts and Contexts 4th Edition Stewart Solutions Manual Visit TestBankDeal.com to get complete for all chapters

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PREFACE

This Complete Solutions Manual contains solutions to all exercises in the texts Single Variable Calculus:Concepts and Contexts, Fourth Edition,and Chapters 1–8 of Calculus:Concepts and Contexts, Fourth Edition,by James Stewart. A student version of this manual is also available; it contains solutions to the odd-numbered exercises in each chapter section,the review sections,the True-False Quizzes,and the Focus on Problem Solving sections,as well as solutions to all the exercises in the Concept Checks. No solutions to the Projects appear in the student version. It is our hope that by browsing through the solutions,professors will save time in determining appropriate assignments for their particular classes.

Some nonstandard notation is used in order to save space. If you see a symbol that you don’t recognize,refer to the Table of Abbreviations and Symbols on page v.

We appreciate feedback concerning errors,solution correctness or style,and manual style. Any comments may be sent directly to us at jeff.cole@anokaramsey.edu or tim@andrew.cmu.edu,or in care of the publisher:Cengage Learning Brooks/Cole,0 Davis Drive,Belmont,CA 94002.

We would like to thank Jim Stewart,for his guidance; Brian Betsill,Kathi Townes,and Rebekah Million,of TECH-arts,for their production services; and Richard Stratton and Jeannine Lawless,of Cengage Learning Brooks/Cole,for entrusting us with this project as well as for their patience and support.

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ABBREVIATIONS AND SYMBOLS

CDconcavedownward

CUconcaveupward

Dthedomainof

FDTFirstDerivativeTest

HAhorizontalasymptote(s)

Iintervalofconvergence

I/DIncreasing/DecreasingTest

IPin ectionpoint(s)

Rradiusofconvergence

VAverticalasymptote(s)

CAS = indicatestheuseofacomputeralgebrasystem.

H = indicatestheuseofl’Hospital’sRule.

= indicatestheuseofFormula intheTableofIntegralsinthebackendpapers.

s = indicatestheuseofthesubstitution { =sin =cos }

c = indicatestheuseofthesubstitution { =cos = sin }.

v
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Australia Brazil Japan Korea Mexico Singapore Spain UnitedKingdom United States DAN CLEGG Palomar College MULTIVARIABLE CALCULUS: CONCEPTS AND CONTEXTS
EDITION ©2010CengageLearning.AllRightsReserved.May notbescanned,copied,orduplicated,orpostedtoapublicly accessiblewebsite,inwholeorinpart. NOT
SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved
COMPLETE SOLUTIONS MANUAL for Stewart’s
FOURTH
FOR

NOT FOR SALE

© 2010 Brooks/Cole, Cengage Learning

ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be reproduced, transmitted, stored, or used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval systems, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without the prior written permission of the publisher.

For product information and technology assistance, contact us at Cengage Learning Customer & Sales Support, 1-800-354-9706

For permission to use material from this text or product, submit all requests online at www.cengage.com/permissions. Further permissions questions can be e-mailed to permissionrequest@cengage.com.

ISBN-13: 978-0-495-56056-2

ISBN-10: 0-495-56056-1

Brooks/Cole

10 Davis Drive Belmont, CA 94002-3098

USA

Cengage Learning is a leading provider of customized learning solutions with office locations around the globe, including Singapore, the United Kingdom, Australia, Mexico, Brazil, and Japan. Locate your local office at www.cengage.com/international.

Cengage Learning products are represented in Canada by Nelson Education, Ltd.

To learn more about Brooks/Cole, visit www.cengage.com/brookscole

Purchase any of our products at your local college store or at our preferred online store www.ichapters.com.

NOTE: UNDER NO CIRCUMSTANCES MAY THIS MATERIAL OR ANY PORTION THEREOF BE SOLD,

OR OTHERWISE REDISTRIBUTED EXCEPT AS MAY BE

Dear Professor or Other Supplement Recipient:

READ IMPORTANT LICENSE INFORMATION

Cengage Learning has provided you with this product (the “Supplement”) for your review and, to the extent that you adopt the associated textbook for use in connection with your course (the “Course”), you and your students who purchase the textbook may use the Supplement as described below. Cengage Learning has established these use limitations in response to concerns raised by authors, professors, and other users regarding the pedagogical problems stemming from unlimited distribution of Supplements.

Cengage Learning hereby grants you a nontransferable license to use the Supplement in connection with the Course, subject to the following conditions. The Supplement is for your personal, noncommercial use only and may not be reproduced, posted electronically or distributed, except that portions of the Supplement may be provided to your students IN PRINT FORM ONLY in connection with your instruction of the Course, so long as such students are advised that they may not copy or distribute any portion of the Supplement to any third

party. You may not sell, license, auction, or otherwise redistribute the Supplement in any form. We ask that you take reasonable steps to protect the Supplement from unauthorized use, reproduction, or distribution. Your use of the Supplement indicates your acceptance of the conditions set forth in this Agreement. If you do not accept these conditions, you must return the Supplement unused within 30 days of receipt.

All rights (including without limitation, copyrights, patents, and trade secrets) in the Supplement are and will remain the sole and exclusive property of Cengage Learning and/or its licensors. The Supplement is furnished by Cengage Learning on an “as is” basis without any warranties, express or implied. This Agreement will be governed by and construed pursuant to the laws of the State of New York, without regard to such State’s conflict of law rules.

Thank you for your assistance in helping to safeguard the integrity of the content contained in this Supplement. We trust you find the Supplement a useful teaching tool.

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PREFACE

This Complete Solutions Manual contains detailed solutions to all exercises in the text Multivariable Calculus:Concepts and Contexts, Fourth Edition (Chapters 8–13 of Calculus: Concepts and Contexts, Fourth Edition) by James Stewart. A Student Solutions Manual is also available,which contains solutions to the odd-numbered exercises in each chapter section,review section,True-False Quiz,and Focus on Problem Solving section as well as all solutions to the Concept Check questions. (It does not,however,include solutions to any of the projects.)

While I have extended every effort to ensure the accuracy of the solutions presented,I would appreciate correspondence regarding any errors that may exist. Other suggestions or comments are also welcome,and can be sent to me at the email address or mailing address below.

I would like to thank James Stewart for entrusting me with the writing of this manual and offering suggestions,Kathi Townes,Stephanie Kuhns,and Rebekah Steele of TECH-arts for typesetting and producing this manual,and Brian Betsill of TECH-arts for creating the illustrations. Brian Karasek prepared solutions for comparison of accuracy and style in addition to proofreading manuscript; his assistance and suggestions were very helpful and much appreciated. Finally, I would like to thank Richard Stratton and Elizabeth Neustaetter of Brooks/Cole,Cengage Learning for their trust,assistance,and patience.

DANCLEGG

dclegg@palomar.edu

Palomar College

Department of Mathematics

1140 West Mission Road

San Marcos,CA 92069

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v CONTENTS 8 ■ INFINITE SEQUENCES AND SERIES 1 8.1 Sequences1 Laboratory Project ■ Logistic Sequences9 8.2 Series13 8.3 The Integral and Comparison Tests; Estimating Sums26 8.4 Other Convergence Tests32 8.5 Power Series39 8.6 Representations of Functions as Power Series46 8.7 Taylor and Maclaurin Series55 Laboratory Project ■ An Elusive Limit69 8.8 Applications of Taylor Polynomials70 Applied Project ■ Radiation from the Stars81 Review83 Focus on Problem Solving95 9 ■ VECTORS AND THE GEOMETRY OF SPACE101 9.1 Three-Dimensional Coordinate Systems101 9.2 Vectors108 9.3 The Dot Product115 9.4 The Cross Product122 Discovery Project ■ The Geometry of a Tetrahedron130 9.5 Equations of Lines and Planes132 Laboratory Project ■ Putting 3D in Perspective141 9.6 Functions and Surfaces143 9.7 Cylindrical and Spherical Coordinates151 Laboratory Project ■ Families of Surfaces156 Review158 Focus on Problem Solving169 ©2010CengageLearning.AllRightsReserved.May notbescanned,copied,orduplicated,orpostedtoapublicly accessiblewebsite,inwholeorinpart. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved

10 ■ VECTOR FUNCTIONS175

10.1 Vector Functions and Space Curves175

10.2 Derivatives and Integrals of Vector Functions185

10.3 Arc Length and Curvature195

10.4 Motion in Space:Velocity and Acceleration208

Applied Project ■ Kepler’s Laws218

10.5 Parametric Surfaces219

Review225

Focus on Problem Solving231

11 ■ PARTIAL DERIVATIVES239

11.1 Functions of Several Variables239

11.2 Limits and Continuity249

11.3 Partial Derivatives256

11.4 Tangent Planes and Linear Approximations272

11.5 The Chain Rule280

11.6 Directional Derivatives and the Gradient Vector290

11.7 Maximum and Minimum Values302

Applied Project ■ Designing a Dumpster318

Discovery Project ■ Quadratic Approximations and Critical Points320

11.8 Lagrange Multipliers323

Applied Project ■ Rocket Science333

Applied Project ■ Hydro-Turbine Optimization335

Review338

Focus on Problem Solving351

12 ■

MULTIPLE INTEGRALS357

12.1 Double Integrals over Rectangles357

12.2 Iterated Integrals362

12.3 Double Integrals over General Regions368

12.4 Double Integrals in Polar Coordinates380

12.5 Applications of Double Integrals386

12.6 Surface Area395

vi ■ CONTENTS
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12.7 Triple Integrals400

Discovery Project ■ Volumes of Hyperspheres416

12.8 Triple Integrals in Cylindrical and Spherical Coordinates417

Applied Project ■ Roller Derby425

Discovery Project ■ The Intersection of Three Cylinders427

12.9 Change of Variables in Multiple Integrals429

Review435

Focus on Problem Solving447 13

VECTOR CALCULUS453

13.1 Vector Fields453

13.2

13.4

13.5

13.6

13.7

13.8

Review505

Focus on Problem Solving515

■ APPENDIXES 519

D Precise Definitions of Limits519

H Polar Coordinates519

Discovery Project ■ Conic Sections in Polar Coordinates543

I Complex Numbers544

CONTENTS ■ vii
Line
Integrals458
13.3 The Fundamental Theorem for Line Integrals466
Green’s Theorem471
Curl and Divergence478
Surface
Integrals486
Stokes’Theorem497
The Divergence Theorem501
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8.1Sequences

1. (a)Asequenceisanorderedlistofnumbers.Itcanalsobedefinedasafunctionwhosedomainisthesetofpositiveintegers.

(b)Theterms approach 8 as becomeslarge.Infact,wecanmake ascloseto 8 aswelikebytaking sufficiently large.

(c)Theterms becomelargeas becomeslarge.Infact,wecanmake aslargeaswelikebytaking sufficientlylarge. 2. (a)FromDefinition1,aconvergentsequenceisasequenceforwhich lim exists.Examples: {1 }, {1 2 }

(b)Adivergentsequenceisasequenceforwhich lim doesnot exist.Examples: { }, {sin }

8
INFINITESEQUENCESANDSERIES
3. Thefirstsixtermsof = 2 +1 are 1 3 , 2 5 , 3 7 , 4 9 , 5 11 , 6 13 .Itappearsthatthesequenceisapproaching 1 2 . lim 2 +1 =lim 1 2+1 = 1 2 4. {cos( 3)}9 =1 = 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 .Thesequencedoesnotappeartohavealimit.Thevalueswillcycle
5. 1 1 3 1 5 1 7 1 9 .Thedenominatorofthe nthtermisthe nthpositiveoddinteger,so = 1 2 1 6. 1 1 3 1 9 1 27 1 81 .Thedenominatorofthe nthtermisthe ( 1)stpowerof3,so = 1 3 1 7. {2 7 12 17 }.Eachtermislargerthantheprecedingoneby 5,so = 1 + ( 1)=2+5( 1)=5 3. 8. 1 4 2 9 3 16 4 25 .Thenumeratorofthe thtermis anditsdenominatoris ( +1)2 .Includingthealternatingsigns, weget =( 1) ( +1)2 . 9. 1 2 3 4 9 8 27 .Eachtermis 2 3 timestheprecedingone,so = 2 3 1 . 10. {5 1 5 1 5 1 }.Theaverageof 5 and 1 is 3,sowecanthinkofthesequenceasalternatelyadding 2 and 2 to 3. Thus, =3+( 1) +1 2 11. = 3+5 2 + 2 = (3+5 2 ) 2 ( + 2 ) 2 = 5+3 2 1+1 ,so 5+0 1+0 =5 as .Converges 12. = 3 3 +1 = 3 3 ( 3 +1) 3 = 1 1+1 3 ,so 1 1+0 =1 as .Converges 13. =1 (0 2) ,so lim =1 0=1 by(7).Converges 1 ©2010CengageLearning.AllRightsReserved.May notbescanned,copied,orduplicated,orpostedtoapublicly accessiblewebsite,inwholeorinpart.
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throughthefirstsixnumbersinthesequence—neverapproachingaparticularnumber.
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2 ¤ CHAPTER8
14. = 3 +1 = 3 ( +1) = 2 1+1 2 ,so as since lim 2 = and lim (1+1 2 )=1.Diverges
lim =lim 1 = lim (1 ) = 0 =1 Converges 16. = 3 +2 5 = 32 3 5 =9 3 5 ,so lim =9lim 3 5 =9 0=0 by(7)with = 3 5 .Converges 17. If = 2 1+8 ,then lim =lim (2 ) (1+8 ) =lim 2 1 +8 = 2 8 = 4 .Since tan iscontinuousat 4 ,by Theorem5, lim tan 2 1+8 =tan lim 2 1+8 =tan 4 =1.Converges
lim =lim +1 9 +1 = lim +1 9 +1 = lim 1+1 9+1 = 1 9 = 1 3 .Converges 19. = ( 1) 1 2 +1 = ( 1) 1 +1 ,so 0 | | = 1 +1 1 0 as ,so 0 bytheSqueezeTheoremand Theorem4.Converges 20. = ( 1) 3 3 +2 2 +1 .Now | | = 3 3 +2 2 +1 = 1 1+ 2 + 1 3 1 as ,butthetermsofthesequence { } alternateinsign,sothesequence 1 3 5 convergesto 1 andthesequence 2 4 6 convergesto +1 Thisshowsthatthegivensequencedivergessinceitstermsdon’tapproachasinglerealnumber. 21. = + 2 1 · = 1+ 2 0 as because 1+ 2 1 and .Converges 22. =cos(2 ).As , 2 0,so cos(2 ) cos0=1 because cos iscontinuous.Converges 23. = 2 = 2 .Since lim 2 H =lim 2 H =lim 2 =0,itfollowsfromTheorem2that lim =0.Converges 24. 2 as ,sosince lim arctan = 2 ,wehave lim arctan2 = 2 .Converges 25. 0 cos 2 2 1 2 [since 0 cos 2 1],sosince lim 1 2 =0, cos 2 2 convergesto 0 bytheSqueezeTheorem. 26. = cos = ( 1) .Since | | = as ,thegivensequencediverges. 27. = 1+ 2 ln = ln 1+ 2 ,so lim ln =lim ln(1+2 ) 1 H =lim 1 1+2 2 2 1 2 =lim 2 1+2 =2 lim 1+ 2 =lim ln = 2 ,sobyTheorem2, lim 1+ 2 = 2 .Convergent ©2010CengageLearning.AllRightsReserved.May notbescanned,copied,orduplicated,orpostedtoapublicly accessiblewebsite,inwholeorinpart. NOT
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INFINITESEQUENCESANDSERIES
15. Becausethenaturalexponentialfunctioniscontinuousat 0,Theorem5enablesustowrite
18. Usingthelastlimitlawforsequencesandthecontinuityofthesquarerootfunction,
FOR
SECTION8.1 SEQUENCES ¤ 3 28. = 21+3 =(21+3 )1 =(21 23 )1 =21 23 =8 21 ,so lim =8lim 21 =8 2lim (1 ) =8 20 =8 byTheorem5,sincethefunction ( )=2 iscontinuousat 0. Convergent 29. = (2 1)! (2 +1)! = (2 1)! (2 +1)(2 )(2 1)! = 1 (2 +1)(2 ) 0 as .Converges 30. = sin2 1+ . | | 1 1+ and lim 1 1+ =0,so 1 1+ 1 1+ lim =0 bythe SqueezeTheorem.Converges 31. {0 1 0 0 1 0 0 0 1 } divergessincethesequencetakesononlytwovalues,0and1,andneverstaysarbitrarilycloseto eitherone(oranyothervalue)for sufficientlylarge. 32. lim (ln )2 H =lim 2(ln )(1 ) 1 =2lim ln H =2lim 1 1 =0,sobyTheorem3, lim (ln )2 =0.Convergent 33. =ln(2 2 +1) ln( 2 +1)=ln 2 2 +1 2 +1 =ln 2+1 2 1+1 2 ln2 as .Convergent 34. 0 | | = 3 ! = 3 1 3 2 3 3 3 ( 1) 3 3 1 3 2 3 [for 2] = 27 2 0 as ,sobytheSqueeze TheoremandTheorem4, {( 3) !} convergesto 0 35. Fromthegraph,itappearsthatthesequenceconvergesto 1 {( 2 ) } convergesto 0 by(7),andhence {1+( 2 ) } convergesto 1+0=1 36. Fromthegraph,itappearsthatthesequenceconvergestoanumber greaterthan 3 lim =lim sin =lim sin =lim 0+ sin = =1 = ©2010CengageLearning.AllRightsReserved.May notbescanned,copied,orduplicated,orpostedtoapublicly accessiblewebsite,inwholeorinpart. NOT FOR SALE SEC INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved
4 ¤ CHAPTER8 INFINITESEQUENCESANDSERIES 37. Fromthegraph,itappearsthatthesequenceconvergesto 1 2 As , = 3+2 2 8 2 + = 3 2 +2 8+1 0+2 8+0 = 1 4 = 1 2 , so lim = 1 2 38. Fromthegraph,itappearsthatthesequenceconvergesto 5 5 = 5 3 +5 5 +5 = 2 5 = 2 · 5 5 as lim 21 =20 =1 Hence, 5 bytheSqueezeTheorem. Alternatesolution: Let =(3 +5 )1 .Then lim ln =lim ln(3 +5 ) H =lim 3 ln3+5 ln5 3 +5 =lim 3 5 ln3+ln5 3 5 +1 =ln5, so lim = ln5 =5,andso 3 +5 convergesto 5 39. Fromthegraph,itappearsthatthesequence { } = 2 cos 1+ 2
provethis,supposethat { } convergesto .If = 2 1+ 2 ,then { } convergesto 1,and lim = 1 = .But =cos ,so lim doesnotexist.Thiscontradictionshowsthat { } diverges. 40. Fromthegraph,itappearsthatthesequenceapproaches 0 0 = 1 · 3 · 5 ····· (2 1) (2 ) = 1 2 · 3 2 · 5 2 ····· 2 1 2 1 2 (1) (1) (1)= 1 2 0 as SobytheSqueezeTheorem, 1 3 5 (2 1) (2 ) convergesto 0. 41. (a) =1000(1 06) 1 =1060, 2 =1123 60, 3 =1191 02, 4 =1262 48,and 5 =1338 23. (b) lim =1000lim (1 06) ,sothesequencedivergesby(7)with =1 06 1. ©2010CengageLearning.AllRightsReserved.May notbescanned,copied,orduplicated,orpostedtoapublicly accessiblewebsite,inwholeorinpart. NOT
EQUENCESANDSERIES AN INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved
is divergent,sinceitoscillatesbetween 1 and 1 (approximately).To
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lim = ,theterms approach as becomeslarge.Becausewecanmake ascloseto aswewish,

SECTION8.1 SEQUENCES ¤ 5 42. (a)Substitute 1 to 6 for in =100 1 0025 1 0 0025 toget 1 =$0, 2 =$0 25, 3 =$0 75, 4 =$1 50, 5 =$2 51,and 6 =$3 76
2 12=24 for toget $70 28
(a)Wearegiventhattheinitialpopulationis 5000,so 0 =5000.Thenumberofcatfishincreasesby 8% permonthandis decreasedby 300 permonth,so 1 = 0 +8% 0 300=1 08 0 300, 2 =1 08 1 300,andsoon.Thus, =1 08 1 300.
0 =5000,weget 1 =5100, 2 =5208, 3 =5325 (roundinganyportionofa catfish), 4 =5451, 5 =5587,and 6 =5734,whichisthenumberofcatfishinthepondaftersixmonths. 44. +1 = 1 2 if isanevennumber 3 +1 if isanoddnumber When 1 =11,thefirst 40 termsare 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4.When 1 =25,thefirst 40 termsare 25, 76, 38, 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4. ThefamousCollatzconjectureisthatthissequencealwaysreaches 1,regardlessofthestartingpoint 1 . 45. (a) 1 =1, +1 =4 for 1. 1 =1, 2 =4 1 =4 1=3, 3 =4 2 =4 3=1, 4 =4 3 =4 1=3, 5 =4 4 =4 3=1.Sincethetermsofthesequencealternatebetween 1 and 3, thesequenceisdivergent. (b) 1 =2, 2 =4 1 =4 2=2, 3 =4 2 =4 2=2.Sinceallofthetermsare 2, lim =2 andhence,the sequenceisconvergent. 46. (a)Since
+1
(b) 1 =1, 2 = 1 1+ 1 = 1 1+1 = 1 2 =0 5, 3 = 1 1+ 2 = 1 1+ 1 2 = 2 3 0 66667, 4 = 1 1+ 3 = 1 1+ 2 3 = 3 5 =0 6, 5 = 1 1+ 4 = 1 1+ 3 5 = 5 8 =0 625, 6 = 1 1+ 5 = 1 1+ 5 8 = 8 13 0 61538, 7 = 1 1+ 6 = 1 1+ 8 13 = 13 21 0 61905, 8 = 1 1+ 7 = 1 1+ 13 21 = 21 34 0 61765, 9 = 1 1+ 8 = 1 1+ 21 34 = 34 55 0 61818, 10 = 1 1+ 9 = 1 1+ 34 55 = 55 89 0 61800.Itappearsthat lim 0 618;hence,thesequenceisconvergent. (c)If =lim then lim +1 = also,so mustsatisfy =1 (1+ ) 2 + 1=0 = 1+ 5 2 0 618 (since hastobenon-negativeifitexists). ©2010CengageLearning.AllRightsReserved.May notbescanned,copied,orduplicated,orpostedtoapublicly accessiblewebsite,inwholeorinpart. NOT FOR SALE SEC INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved
(b)Fortwoyears,use
43.
(b)Usingtherecursiveformulawith
willalsobeclose,andso lim +1 =

CHAPTER8 INFINITESEQUENCESANDSERIES

47. (a)Let bethenumberofrabbitpairsinthe nthmonth.Clearly 1 =1= 2 .Inthe nthmonth,eachpairthatis 2 ormoremonthsold(thatis, 2 pairs)willproduceanewpairtoaddtothe 1

= ( 1) alternateinsign,sothesequenceisnotmonotonic.Thefirstfivetermsare

lim | | =lim = ,thesequenceisnotbounded.

52. = + 1 definesanincreasingsequencesincethefunction ( )= + 1 isincreasingfor

for 1.]Thesequenceisunboundedsince as .(Itis,however,boundedbelowby 1 =2.)

.

53. Since { } isadecreasingsequence, +1 forall 1.Becauseallofitstermsliebetween 5 and 8, { } isa boundedsequence.BytheMonotonicSequenceTheorem, { } isconvergent;thatis, { } hasalimit mustbelessthan 8 since { } isdecreasing,so 5 8

54. (a)Let bethestatementthat +1 and 3 1 isobviouslytrue.Wewillassumethat istrueand thenshowthatasaconsequence +1 mustalsobetrue. +2 +1 2+ +1 2+

6 ¤
= 1 + 2 ,sothat
(b) = +1 1 = 1 = 1 + 2 1 =1+ 2 1 =1+ 1 1 / 2 =1+ 1 2 .If =lim , then =lim 1 and =lim 2 ,so mustsatisfy =1+ 1 2 1=0 = 1+ 5 2 [since mustbepositive]. 48. For 2, 2 2, 2 2 2, , 1 =21 2 , 2 =23 4 , 3 =27 8 , ,so =2(2 1) 2 =21 (1 2 ) . lim =lim 21 (1 2 ) =21 =2 Alternatesolution:Let =lim .(Wecouldshowthelimitexistsbyshowingthat { } isboundedandincreasing.) Then mustsatisfy = 2 2 =2 ( 2)=0. =0 sincethesequenceincreases,so =2. 49. = 1 2 +3 isdecreasingsince +1 = 1 2( +1)+3 = 1 2 +5 1 2 +3 = foreach 1.Thesequenceis boundedsince 0 1 5 forall 1.Notethat 1 = 1 5 50. = 2 3 3 +4 definesanincreasingsequencesincefor ( )= 2 3 3 +4 , 0 ( )= (3 +4)(2) (2 3)(3) (3 +4)2 = 17 (3 +4)2 0.Thesequenceisboundedsince 1 = 1 7 for 1, and 2 3 3 2 3 = 2 3 for 1. 51. Thetermsof
1, 2, 3, 4,and 5
pairsalreadypresent.Thus,
{ } = { },theFibonaccisequence.
Since
1 [ 0 ( )=1 1 2 0
©2010CengageLearning.AllRightsReserved.May notbescanned,copied,orduplicated,orpostedtoapublicly accessiblewebsite,inwholeorinpart. NOT
EQUENCESANDSERIES AN INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved
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SECTION8.1 SEQUENCES ¤ 7 2+ +1 2+ +1 ,whichistheinductionhypothesis. +1 3 2+ 3 2+ 9 7,whichiscertainlytruebecauseweareassumingthat 3.So istrueforall ,andso 1 3 (showingthatthesequenceisbounded),andhencebytheMonotonicSequenceTheorem, lim exists. (b)If =lim ,then lim +1 = also,so = 2+ 2 =2+ 2 2=0 ( +1)( 2)=0 =2 [since can’tbenegative]. 55. 1 =1, +1 =3 1 .Weshowbyinductionthat { } isincreasingandboundedaboveby 3.Let betheproposition that +1 and 0 3.Clearly 1 istrue.Assumethat istrue.Then +1 1 +1 1 1 +1 1 .Now +2 =3 1 +1 3 1 = +1 +1 .Thisprovesthat { } isincreasingandbounded aboveby 3,so 1= 1 3,thatis, { } isbounded,andhenceconvergentbytheMonotonicSequenceTheorem. If =lim ,then lim +1 = also,so mustsatisfy =3 1 2 3 +1=0 = 3 ± 5 2 . But 1,so = 3+ 5 2 . 56. 1 =2, +1 = 1 3 .Weuseinduction.Let bethestatementthat 0 +1 2.Clearly 1 istrue,since 2 =1 (3 2)=1.Nowassumethat istrue.Then +1 +1 3 +1 3 +2 = 1 3 +1 1 3 = +1 .Also +2 0 [since 3 +1 ispositive]and +1 2 bytheinduction hypothesis,so +1 istrue.Tofindthelimit,weusethefactthat lim =lim +1 = 1 3 2 3 +1=0 = 3 ± 5 2 .But 2,sowemusthave = 3 5 2 . 57. (0 8) 0 000001 ln(0 8) ln(0 000001) ln(0 8) ln(0 000001) ln(0 000001) ln(0 8) 61 9,so mustbeatleast 62 tosatisfythegiveninequality. 58. (a)If iscontinuous,then ( )= lim =lim ( )=lim +1 =lim = byExercise46(a). (b)Byrepeatedlypressingthecosinekeyonthecalculator(thatis,takingcosineofthepreviousanswer)untilthedisplayed valuestabilizes,weseethat 0 73909 59. (a)Suppose { } convergesto .Then +1 = + lim +1 = lim +lim = + 2 + = ( + )=0 =0 or = . (b) +1 = + = 1+ since 1+ 1 ©2010CengageLearning.AllRightsReserved.May notbescanned,copied,orduplicated,orpostedtoapublicly accessiblewebsite,inwholeorinpart. SEC INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved Single Variable Calculus Concepts and Contexts 4th Edition Stewart Solutions Manual Visit TestBankDeal.com to get complete for all chapters

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