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# AMS 510 - AMS-510 Analytical Method for AMS Final Exam You...

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AMS-510 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

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(2). Using L’Hospital’s rule, find the following limit lim x →∞ R x 0 e t 2 dt 2 R x 0 e 2 t 2 dt Solution: lim x →∞ R x 0 e t 2 dt 2 R x 0 e 2 t 2 dt L 0 H = lim x →∞ 2 R x 0 e t 2 dt · e x 2 e 2 x 2 = lim x →∞ 2 R x 0 e t 2 dt e x 2 L 0 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

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(4). Find the following limit: lim n →∞ 1 n ln 1 + 1 n + ln 1 + 2 n + · · · ln 1 + n n Solution: lim n →∞ 1 n ln 1 + 1 n + ln 1 + 2 n + · · · ln 1 + n n = Z 1 0 ln(1 + x ) dx = x ln(1 + x ) | 1 x =0 - Z 1 0 x · 1 1 + x dx = ln 2 - Z 1 0 1 - 1 1 + x dx = ln 2 - [ x - ln(1 + x )] 1 x
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