Gilbert_ReviewFinalsol

Gilbert_ReviewFinalsol - moseley (cmm3869) – FinalREVIEW...

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Unformatted text preview: moseley (cmm3869) – FinalREVIEW – Gilbert – (56380) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the value of lim x → 2 ln( x 2 − 3) 4 x − 8 . 1. limit = 5 4 2. limit does not exist 3. limit = 1 correct 4. limit = 1 4 5. limit = 1 2 6. limit = 3 2 Explanation: The limit in question is of the form lim x → 2 f ( x ) g ( x ) where f, g are differentiable functions and lim x → 2 f ( x ) = 0 , lim x → 2 g ( x ) = 0 . L’Hospital’s Rule can thus be applied: lim x → 2 f ( x ) g ( x ) = lim x → 2 f ′ ( x ) g ′ ( x ) . Now f ′ ( x ) = 2 x x 2 − 3 , g ′ ( x ) = 4 . In this case, therefore, lim x → 2 ln( x 2 − 3) 4 x − 8 = 1 . 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 is divergent 2. I = 1 2 π 2 correct 3. I = 1 4 4. I = 1 4 π 2 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 . moseley (cmm3869) – FinalREVIEW – Gilbert – (56380) 2 Now lim t → 1 − (sin − 1 t ) 2 = π 2 4 . Consequently, I is convergent and I = 1 2 π 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 2. converges with limit = 0 correct 3. converges with limit = 1 4. converges with limit = 4 5. converges with limit = 1 4 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)! = 1 2 n (2 n + 1) . Consequently, the given sequence converges with limit = 0 . 004 10.0 points Let f be a continuous, positive, decreasing function on [5 , ∞ ). Compare the values of the integral A = integraldisplay 18 5 f ( z ) dz and the series B = 18 summationdisplay n = 6 f ( n ) . 1. A = B 2. A > B correct 3. A < B Explanation: In the figure 5 6 7 8 9 . . ....
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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.

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Gilbert_ReviewFinalsol - moseley (cmm3869) – FinalREVIEW...

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