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Unformatted text preview: AMS510 Analytical Method for AMS Midterm Exam 1 Do all the problems, six right will get you full credit (1). Find the following limits (a). lim n →∞ n + 4 n n (b). lim n →∞ 1 2 n + 1 3 n 1 n (c). lim x → cos 2 x 1 x 4 x 2 (d). lim n →∞ 1 n 1 + 1 n 2 + 1 + 2 n 2 + ··· 1 + 2 n n 2 ! Solution: (a). L = lim n →∞ n + 4 n n = lim n →∞ 1 + 4 n n Let n = 4 x , L = lim n →∞ 1 + 1 x 4 x = lim n →∞ 1 + 1 x x 4 = e 4 (b). lim n →∞ 1 2 n 1 n ≤ lim n →∞ 1 2 n + 1 3 n 1 n ≤ lim n →∞ 1 2 n + 1 2 n 1 n LHS = 1 2 , RHS = lim n →∞ 2 1 2 n 1 n = 1 2 lim n →∞ 2 1 n = 1 2 Therefore the limit is 1 2 . 1 (c). L = lim x → cos 2 x 1 x 4 x 2 Using L’Hospital’s rule, we have L = lim x → 2 sin 2 x 4 x 3 2 x = lim x → 4 cos 2 x 12 x 2 2 = 4 2 = 2 (d). L = lim n →∞ 1 n 1 + 1 n 2 + 1 + 2 n 2 + ··· 1 + 2 n n 2 ! This the limit of Riemann sum L = Z 3 1 x 2 dx = 1 3 x 3  3 1 = 9 1 3 = 26 3 (2). Given function f ( x ) =  x  3 , find:...
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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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