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Unformatted text preview: rhodes (ajr2283) FinalREVIEW 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 Find the value of lim x 4 ln( x 2 15) 5 x 20 . 1. limit = 4 5 2. limit = 2 3. limit = 8 5 correct 4. limit = 9 5 5. limit = 2 5 6. limit does not exist Explanation: The limit in question is of the form lim x 4 f ( x ) g ( x ) where f, g are differentiable functions and lim x 4 f ( x ) = 0 , lim x 4 g ( x ) = 0 . LHospitals Rule can thus be applied: lim x 4 f ( x ) g ( x ) = lim x 4 f ( x ) g ( x ) . Now f ( x ) = 2 x x 2 15 , g ( x ) = 5 . In this case, therefore, lim x 4 ln( x 2 15) 5 x 20 = 8 5 . 002 10.0 points Determine if the improper integral I = integraldisplay 1 4 sin 1 x 1 x 2 dx is convergent or divergent, and if convergent, find its value. 1. I = 1 4 2. I is divergent 3. I = 1 4 2 4. I = 1 2 2 correct 5. I = 1 2 Explanation: The integral is improper because lim x 1 sin 1 x 1 x 2 = . Thus we have to check if lim t 1 integraldisplay t 4 sin 1 x 1 x 2 dx is convergent or divergent. To determine integraldisplay 4 sin 1 x 1 x 2 dx set u = sin 1 x . Then du = 1 1 x 2 dx, and so integraldisplay 4 sin 1 x 1 x 2 dx = 4 integraldisplay udu = 2 u 2 + C , from which it follows that integraldisplay t 4 sin 1 x 1 x 2 dx = 2 bracketleftBig (sin 1 x ) 2 bracketrightBig t = 2(sin 1 t ) 2 . rhodes (ajr2283) FinalREVIEW Gilbert (56380) 2 Now lim t 1 (sin 1 t ) 2 = 2 4 . Consequently, I is convergent and I = 1 2 2 . 003 10.0 points Determine if the sequence { a n } converges when a n = (2 n + 1)! (2 n 1)! , and if it converges, find the limit. 1. does not converge correct 2. converges with limit = 1 3. converges with limit = 1 4 4. converges with limit = 4 5. converges with limit = 0 Explanation: By definition, m ! is the product m ! = 1 . 2 . 3 . . . . .m of the first m positive integers. When m = 2 n 1, therefore, (2 n 1)! = 1 . 2 . 3 . . . . (2 n 1) , while (2 n + 1)! = 1 . 2 . 3 . . . . . (2 n 1)2 n (2 n + 1) . when m = 2 n + 1. But then, (2 n + 1)! (2 n 1)! = 2 n (2 n + 1) . Consequently, the given sequence does not converge . 004 10.0 points Let f be a continuous, positive, decreasing function on [2 , ). Compare the values of the integral A = integraldisplay 16 2 f ( x ) dx and the series B = 16 summationdisplay n = 3 f ( n ) . 1. A > B correct 2. A < B 3. A = B Explanation: In the figure 2 3 4 5 6 . . ....
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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 at Austin.
 Spring '07
 TextbookAnswers
 Multivariable Calculus

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