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Gilbert_ReviewFinalsol

Gilbert_ReviewFinalsol - moseley(cmm3869 FinalREVIEW...

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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 0 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 0 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 u du = 2 u 2 + C , from which it follows that integraldisplay t 0 4 sin 1 x 1 x 2 dx = 2 bracketleftBig (sin 1 x ) 2 bracketrightBig t 0 = 2(sin 1 t ) 2 .
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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 . . . a 6 a 7 a 8 a 9 the bold line is the graph of f on [5 , ) and the areas of the rectangles the terms in the series summationdisplay n =6 a n , a n = f ( n ) . Clearly from this figure we see that a 6 = f (6) < integraldisplay 6 5 f ( z ) dz, a 7 = f (7) < integraldisplay 7 6 f ( z ) dz ,
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moseley (cmm3869) – FinalREVIEW – Gilbert – (56380) 3 while a 8 = f (8) < integraldisplay 8 7 f ( z ) dz, a 9 = f (9) < integraldisplay 9 8 f ( z ) dz , and so on. Consequently, A > B .
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