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Unformatted text preview: . Solution #1: Since lim n →∞ 12 n 2 + 2 n 5 n 5 + 10 n + 1 is of the form “ ∞ ∞ ”, we may use l’Hopital’s rule (repeatedly) to ﬁnd lim n →∞ 12 n 2 + 2 n 5 n 5 + 10 n + 1 = lim n →∞ 24 n + 10 n 4 5 n 4 + 10 = lim n →∞ 24 + 40 n 3 20 n 3 = lim n →∞ 120 n 2 60 n 2 = lim n →∞ 240 n 120 n = lim n →∞ 240 120 = 2 , so that lim n →∞ a n = 2. Solution #2: The largest power in the numerator and denominator of a n is n 5 , so if we divide numerator and denominator by n 5 we ﬁnd lim n →∞ 12 n 2 + 2 n 5 n 5 + 10 n + 1 = lim n →∞ (12 n 2 + 2 n 5 ) ÷ n 5 ( n 5 + 10 n + 1) ÷ n 5 = lim n →∞ 12 n 3 + 2 1 + 10 n 4 + 1 n 5 = 0 + 2 1 + 0 + 0 = 2 , leading once again to the conclusion that lim n →∞ a n = 2. 2...
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This note was uploaded on 04/02/2012 for the course MTH 132 taught by Professor Kihyunhyun during the Fall '10 term at Michigan State University.
 Fall '10
 KIHYUNHYUN
 Calculus

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