Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

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46 D CHAPTER 11 INFINITE SEQUENCES AND SERIES 19. 3n n an= 1+ 6n 1 0.4286 2 0.4615 3 0.4737 ' 4 0.4800 5 0.4839 6 0.4865 7 0.4884 8 0.4898 9 0.4909 10 0.4918 0.5 0.4 It appears that lim an = 0.5. n-+oo 0 . . . . . . . . . . 5 10 n lim ~ = lim ( 3 n)jn = lim 3 = ~ = .! n-+oo 1 + 6n n-+oo (1 + 6n)jn n-+oo 1/n + 6 6 2 21 . n an= 1 + (-U~ 1 0.5000 2 1.2500 3 0.8750 4 1.0625 5 0.9688 6 1.0156 I 7 0.9922 8 1.0039 9 0.9980 10 1.0010 I It appears that lim a,. = 1. n-co . . . . . 0 5 10 n lim (1+(- ~f) =lim 1+ lim (- ~f=1+0 = 1 s ince n -oo n-too n-oo "" 23. an = 1 - (0.2)", so lim a,. = 1 - 0 = 1 by (9) . Converges n - oo 3 + 5n 2 (3 + 5n 2 )jn 2 5 3/n 2 5 + 0 25. a,, = n + n2 = (n + n2)/n2 = 1 + l /n, so a,, -+ i + 0 = 5 as n -+ oo. Converges 27.. Because the natural exponential function is continuous at 0, Theorem 7 enables us to write lim an= lim e 1 fn = e 1 imn-oo(l /nl = e 0 = 1. Converges ,.,_ex:> n-oo 2mr h lim' b 1' (2mr)/n lim 271' 271' 71' s· . . .,. b 1 + n n-+oo n-+oo + n n n->oo 1 n + 29. I fb n = ---, t en n = 1m ( )/ = / 8 1 8 2 Theorem 7, lim tan ( n 71' 1 8 ) = tan ( lim 2 n1r ) =tan~ = 1. Converges n --->oo + n n-+oo 1 + 8 n 4 8 = - 8 = - 4 . mce tan IS contmuous at 4 , y n2 n2 j,;r;J .,jii 31. an = = = , so an -+ oo as n--+ oo since lim .,jii = oo and vn 3 + 4n ../n3 + 4n/v'n3 Jl + 4jn2 n--->OO lim j l + 4/ n 2 = 1. 1\-+00 Diverges 1 12 = - 1 (0) = 0, so lim an = 0 by (6). Converges n--+oo n-+oo 2vn n-+oo n 2 n-oo 33. lim Jan! = lim I (-%"I = - 2 1 lim 1 ® 2012 Cengage Lcaminl,;. All Rights ReserYt.-d. Mny not be scanned. coried, or duplicated, or posted ton publicly accessible website, in \vholc or in pan.
SECTION 11.1 SEQUENCES 0 47 35. a,. = cos( n /2). This sequence diverges since the tenns don't approach any particular real number as n---> oo. I The terms take on val4es between - 1 and I. _ (2n - 1)! = (2n - 1}! · = 1 · ___. 0 as n---> oo. C 37· an - (2n + 1)! (2n + 1}(2n}(2n- 1)! (2n + 1}(2n) onverges 1 + e - 2n --'------ ---> 0 as n---> oo because 1 + e- 2 n ---> 1 and e" - e-n -+ oo. en- e n Converges 2 -n n 2 • • x 2 H 2x 11 2 41. an = n e = -. Smce lim - = lim - = lim - = 0, it follows from Theorem 3 that lim an= 0. Converges en X--+00 e X X-tCO eX X--tOO e:t n-oo 2 43. 0 < cos n < _..!:.._ · [since 0 :s; cos 2 1 n :s; 1], so since lim 2 - 2n - 2" n->oe> " n · = 0, { 'c 0 82 n } converges to 0 by the Squeeze Theorem. 2 . ( / ) sin(1/ / n n) . s· mce li m sin(1/x) / = 1 . tm sin t [ w h ere t = 1 x = 1, tt •O ows · om Theorem 3 45. an = n sm 1 n = 1 that {an } converges to I . x ..... oo 1 X t-o+ t / ] . ., II fr ( 2)"' . ( 2)" 2 = lim = 2 ::::} x ->oo 1 + 2/x lim 1 + - = lim e 1 n Y = e 2 , so by Theorem 3, lim 1 + - = e 2 • Converges !t-oo X :r-oo n -+oo n ( 2n2 1) (2 1/n 2 ) 49. an = In(2n 2 + 1)- ln(n 2 + 1) =In n 2 + = In . + / n 1 1 1 2 -+ In 2 as n ---> oo. Converges 51. a.,, = arctan(ln n). Let f(x) = arctan(ln x). Then lim f(x) = ~s ince In x -+ oo as x---> oo and arctan is continuous. x - oo Thus, lim a,= lim f(n) = ~- n-+oo n -+oo Converges 53. {0, 1, 0, 0, 1, 0, 0, 0, 1, ... } diverges since the sequence takes on only two values, 0 and I, and never stays arbitrarily close to either one (or any other value) for n sufficiently large. n ! 1 2 3 (n- 1) n 1 n n 55. an= 2 n = 2 · 2 · 2 · · · · · - - 2 - · 2 2': 2 · 2 (for n > 1] = 4--> oo as n--+ oo, so {an} diverges. 57. 2~------------~ From the graph, it appears.that the sequence converges to 1. { ( -2/e)n} converges to 0 by (7), and hence {1 + ( - 2/e)"} .. . . .. . . converges to 1 + 0 = 1. 0 '---~--~--~---~ 21 ® 2012 Ccng;,gc Lc011T1ing. AU RighiS R.escn·l.-d. May not be scanned, copied. or duplicated. or posted to a publicly accessible website, in whole or in part.
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- STUDENT SOLUTIONS MANUAL for STEW
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CHAPTER 11 REVIEW 0 97 J'(xn)(xn -
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CHAPTER 11 REVIEW 0 99 oo f (n) (O)
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CHAPTER 11 REVIEW 0 101 l 23. Consi
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49./- 1 - dx = -ln{4- x) + C and 4-
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D PROBLEMS PLUS 1. It would be far
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CHAPTER 11 PROBLEMS PLUS D 107 l 9.
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CHAPTER 11 PROBLEMS PLUS 0 109 x x
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112 0 CHAPTER 12 VECTORS AND THE GE
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D PROBLEMS PLUS 1. Since three-dime
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l CHAPTER 12 PROBLEMS PLUS 0 149 Eq
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13 D VECTOR FUNCTIONS 13.1 Vector F
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SECTION 13.1 VECTOR FUNCTIONS AND S
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SECTION 13.1 VECTOR FUNCTIONS AND S
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SECTION 13.2 DERIVATIVES AND INTEGR
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SECTION 13.3 ARC LENGTH AND CURVATU
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SECTION 13.4 MOTION IN SPACE: VELOC
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CHAPTER 13 REVIEW D 173 43. The tan
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5. f~(t 2 i +tcos 1rtj +sin 1rt k )
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CHAPTER 13 REVIEW 0 177 23. (a) r (
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184 0 CHAPTER 14 PARTIAL DERIVATIVE
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D PROBLEMS PLUS 1. The areas of the
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15 D MULTIPLE INTEGRALS 15.1 Double
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SECTION 15.2 ITERATED INTEGRALS D 2
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l SECTION 15.3 DOUBLE INTEGRALS OVE
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SECTION 15.3 DOUBLE INTEGRALS OVER
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SECTION 15.3 DOUBLE INTEGRALS OVER
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61 . Since m :S j(x, y) :S M, ffD m
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SECTION 15.4 DOUBLE INTEGRALS IN PO
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SECTION 15.5 APPLICATIONS OF DOUBLE
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-I SECTION 15.6 SURFACE AREA 0 267
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SECTION 15.7 TRIPLE INTEGRALS 0 269
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Therefore E = { (x, y, z) I -2 ~ X~
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43. I,. = foL foL foL k(y2 + z2)dz.
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SECTION 15.8 TRIPLE INTEGRALS IN CY
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M xv = I~1f I: I:2 6 - 3 r 2 (zK) r
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SECTION 15.9 TRIPLE INTEGRALS IN SP
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(b) The wedge in question is the sh
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SECTION 15.10 CHANGE OF VARIABLES I
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CHAPTER 15 REVIEW 0 289 15 Review C
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CHAPTER 15 REVIEW D 293 33. Using t
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8 8 8 (b)clivF = 'V ·iF=- (x+yz) +
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that is, D = {( x, y) I x 2 + y 2 :
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SECTION 16.7 SURFACE INTEGRALS 0 32
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dS SECTION 16.9 THE DIVERGENCE THEO
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CHAPTER 16 REVIEW 0 337 27. JI 5 cu
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CHAPTER 16 REVIEW 0 339 TRUE-FALSE
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CHAPTER 16 REVIEW D 341 Alternate s
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3.c6 0 CHAPTER 17 SECOND-ORDER DIFF
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0 APPENDIX Appendix H Complex Numbe
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APPENDIX H COMPLEX NUMBERS 0 361 43
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Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

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