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Unformatted text preview: Version 037 – 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 3 x − 1 sin2 x exists, and if it does, find its value. 1. limit = 0 2. limit = 2 3 3. limit = −∞ 4. none of the other answers 5. limit = ∞ 6. limit = 3 2 correct Explanation: Set f ( x ) = e 3 x − 1 , g ( x ) = sin 2 x. Then f and g are differentiable functions such that lim x → f ( x ) = 0 , lim x → g ( x ) = 0 . Thus L’Hospital’s Rule can be applied: lim x → f ( x ) g ( x ) = lim x → f ′ ( x ) g ′ ( x ) . But f ′ ( x ) = 3 e 3 x , g ′ ( x ) = 2 cos 2 x, and so lim x → f ′ ( x ) = 3 , lim x → g ′ ( x ) = 2 . Consequently, the limit exists and limit = 3 2 . 002 10.0 points Determine if the improper integral I = integraldisplay ∞ 1 tan − 1 x 1 + x 2 dx converges, and if it does, find its value. 1. I = 3 16 π 2 2. I = 3 32 3. I = 1 8 π 2 4. I = 3 32 π 2 correct 5. I = 1 8 6. I does not converge 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 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 = π 4 , and x = t = ⇒ u = tan − 1 t . In this case I t = integraldisplay tan 1 t π/ 4 udu = 1 2 bracketleftBig u 2 bracketrightBig tan 1 t π/ 4 . Version 037 – FINAL – Gilbert – (56380) 2 Thus I t = 1 2 braceleftBig ( tan − 1 t ) 2 − π 2 16 bracerightBig . On the other hand, lim t →∞ tan − 1 t = π 2 . Consequently, I converges and has value I = 1 2 parenleftBig 1 4 − 1 16 parenrightBig π 2 = 3 32 π 2 . 003 10.0 points Determine whether the sequence { a n } con verges or diverges when a n = ln(4 n 3 ) ln(3 n 4 ) , and if it converges, find the limit. 1. converges with limit = 4 3 2. converges with limit = 0 3. diverges 4. converges with limit = ln 4 ln 3 5. converges with limit = 3 4 correct Explanation: By properties of logs, ln(4 n 3 ) = ln 4 + 3 ln n , ln(3 n 4 ) = ln 3 + 4 ln n . Thus a n = ln4 + 3 ln n ln3 + 4 ln n = 3 + ln 4 ln n 4 + ln 3 ln n . On the other hand, lim n →∞ ln 4 ln n = lim n →∞ ln 3 ln n = 0 . Properties of limits thus ensure that the given sequence converges with limit = 3 4 . 004 10.0 points Which, if any, of the following series con verge? A . ∞ summationdisplay n =1 1 1 + n 2 B . 1 + 1 4 + 1 9 + 1 16 + . . . 1. B only 2. both of them correct 3. A only 4. neither of them Explanation: A. Use f ( x ) = 1 1 + x 2 . Then integraldisplay ∞ 1 f ( x ) dx is convergent (tan − 1 integral), so series converges....
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This note was uploaded on 09/27/2010 for the course M 408 D taught by Professor Textbookanswers during the Spring '07 term at University of Texas.
 Spring '07
 TextbookAnswers
 Multivariable Calculus

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