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Unformatted text preview: AMS510 Analytical Method for AMS Final Exam You are encouraged to do all the problems, the best 12 will be counted (1). Find the limit lim n →∞ ( √ n + 2 2 √ n + 1 + √ n ) Solution: lim n →∞ √ n + 2 2 √ n + 1 + √ n = lim n →∞ [( √ n + 2 √ n + 1) ( √ n + 1 √ n )] = lim n →∞ 1 √ n + 2 + √ n + 1 1 √ n + 1 + √ n = lim n →∞ ( √ n + 1 + √ n ) ( √ n + 2 + √ n + 1) ( √ n + 2 + √ n + 1)( √ n + 1 + √ n ) = lim n →∞ √ n √ n + 2 ( √ n + 2 + √ n + 1)( √ n + 1 + √ n ) = lim n →∞ 2 ( √ n + 2 + √ n + 1)( √ n + 1 + √ n )( √ n + √ n + 2) = 0 1 (2). Using L’Hospital’s rule, find the following limit lim x →∞ R x e t 2 dt 2 R x e 2 t 2 dt Solution: lim x →∞ R x e t 2 dt 2 R x e 2 t 2 dt L H = lim x →∞ 2 R x e t 2 dt · e x 2 e 2 x 2 = lim x →∞ 2 R x e t 2 dt e x 2 L H = lim x →∞ 2 e x 2 e x 2 · 2 x = lim x →∞ 1 x = 0 2 (3). Evaluate the integral Z x 2 + 1 ( x + 1) 2 ( x 1) dx Solution: Put x 2 + 1 ( x + 1) 2 ( x 1) = a x + 1 + b ( x + 1) 2 + c x 1 and we can get x 2 + 1 = a ( x + 1)( x 1) + b ( x 1) + c ( x + 1) 2 Comparing the coefficient gives a = 1 2 ,b = 1 ,c = 1 2 Then Z x 2 + 1 ( x + 1) 2 ( x 1) dx = Z 1 2 1 x + 1 1 ( x + 1) 2 + 1 2 1 x 1 dx = 1 2 ln  x + 1  + 1 x + 1 + 1 2 ln  x 1  + c = 1 2 ln  x 2 1  + 1 x + 1 + c 3 (4). Find the following limit:(4)....
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This note was uploaded on 02/28/2012 for the course AMS 510 taught by Professor Feinberg,e during the Fall '08 term at SUNY Stony Brook.
 Fall '08
 Feinberg,E

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