11.1.25-eg - Solution#1 Since lim n →∞ 12 n 2 2 n 5 n 5...

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Example Consider the sequence { a n } where the nth term is given by a n = 12 n 2 + 2 n 5 n 5 + 10 n + 1 Determine if the sequence converges or diverges. If the sequence converges, find its limit. Note that the nth term a n is the same as the value of the function f ( x ) = 12 x 2 + 2 x 5 x 5 + 10 x + 1 evaluated at the point x = n . So the points ( n,a n ) = ( n,f ( n )) lie on the graph of the function y = f ( x ) in the x - y plane, and moreover, lim n →∞ a n = lim n →∞ f ( n ) = lim x →∞ f ( x ) . Thus, any techniques that can be used for a limit of the form lim x →∞ f ( x ) can be used for the limit lim n →∞ a n of the sequence { a n } . For the function f ( x ) given above, we can find lim x →∞ f ( x ) using l’Hˆ opital’s rule, and also by dividing numerator and denominator of f ( x ) by the largest power of x appearing in both numerator and denominator. This leads to two different approaches to the problem of deter- mining the limit lim n →∞ a n of the sequence { a n }
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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.

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11.1.25-eg - Solution#1 Since lim n →∞ 12 n 2 2 n 5 n 5...

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