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Unformatted text preview: Version 067 FINAL Gilbert (56380) 1 This printout should have 25 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Determine if lim x e 5 x 1 sin3 x exists, and if it does, find its value. 1. limit = 2. none of the other answers 3. limit = 5 3 correct 4. limit = 0 5. limit = 6. limit = 3 5 Explanation: Set f ( x ) = e 5 x 1 , g ( x ) = sin 3 x. Then f and g are differentiable functions such that lim x f ( x ) = 0 , lim x g ( x ) = 0 . Thus LHospitals Rule can be applied: lim x f ( x ) g ( x ) = lim x f ( x ) g ( x ) . But f ( x ) = 5 e 5 x , g ( x ) = 3 cos 3 x, and so lim x f ( x ) = 5 , lim x g ( x ) = 3 . Consequently, the limit exists and limit = 5 3 . 002 10.0 points Determine if the improper integral I = integraldisplay 1 6 tan 1 x 1 + x 2 dx converges, and if it does, find its value. 1. I = 9 8 2 2. I does not converge 3. I = 3 4 4. I = 9 16 2 correct 5. I = 9 16 6. I = 3 4 2 Explanation: The integral I is improper because the in terval of integration is infinite. Thus the integral will converge if lim t I t , I t = integraldisplay t 1 6 tan 1 x 1 + x 2 dx, exists, and its value will then be the value of the limit. The substitution u = tan 1 x is suggested. For then du = 1 1 + x 2 dx, while x = 1 = u = 4 , and x = t = u = tan 1 t . In this case I t = integraldisplay tan 1 t / 4 6 udu = 3 bracketleftBig u 2 bracketrightBig tan 1 t / 4 . Version 067 FINAL Gilbert (56380) 2 Thus I t = 3 braceleftBig ( tan 1 t ) 2 2 16 bracerightBig . On the other hand, lim t tan 1 t = 2 . Consequently, I converges and has value I = 3 parenleftBig 1 4 1 16 parenrightBig 2 = 9 16 2 . 003 10.0 points Determine whether the sequence { a n } con verges or diverges when a n = ln(5 n 6 ) ln(4 n 3 ) , and if it converges, find the limit. 1. converges with limit = 2 correct 2. converges with limit = 5 4 3. diverges 4. converges with limit = ln 5 ln 4 5. converges with limit = 0 Explanation: By properties of logs, ln(5 n 6 ) = ln 5 + 6 ln n , ln(4 n 3 ) = ln 4 + 3 ln n . Thus a n = ln5 + 6 ln n ln4 + 3 ln n = 6 + ln 5 ln n 3 + ln 4 ln n . On the other hand, lim n ln 5 ln n = lim n ln 4 ln n = 0 . Properties of limits thus ensure that the given sequence converges with limit = 2 . 004 10.0 points Which, if any, of the following series con verge? A . summationdisplay n =1 n 1 + n 2 B . 1 + 1 4 + 1 9 + 1 16 + . . . 1. both of them 2. neither of them 3. B only correct 4. A only Explanation: A. Use f ( x ) = x 1 + x 2 . Then integraldisplay 1 f ( x ) dx is divergent (ln integral), so series di verges....
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This note was uploaded on 04/12/2011 for the course M 408d taught by Professor Sadler during the Spring '07 term at University of Texas at Austin.
 Spring '07
 Sadler
 Multivariable Calculus

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