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Hwk4_Solns

# Hwk4_Solns - Math 3110 Spring 09 Homework 4 Selected...

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Math 3110 Spring ’09 Homework 4: Selected Solutions [5.1-1 (b)]: Claim : lim n →∞ n - 3 2 n - 1 ( 2 - n ) = 0 Proof Use linearity, then the quotient rule and finally the product rule for limits: lim n →∞ n - 3 2 n - 1 ( 2 - n ) = lim n →∞ 1 n 1 n n - 3 2 n - 1 ( 2 - n ) = lim n →∞ 1 - 3 n 2 - 1 n ( 2 - n ) = 0 [5.1-4]: Claim : If a n b n L , b n 6 = 0 for all n and b n 0, the a n 0. Proof There were two main ways that students tried to prove this - the first, using the product rule, is very easy whereas the second, using the quotient rule, is a little more involved. I’ll show both versions below. Product Rule Version: Let c n = a n b n . We are given that both c n and b n converge to finite limits, so by the product rule lim n →∞ c n b n = lim c n lim n →∞ b n = L · 0 = 0 . And, of course, c n b n = a n b n · b n = a n , so lim n →∞ a n = 0 Quotient Rule Version: Using the quotient rule directly (without employing the product rule as above) means that we can no longer assume that a n converges at all. The natural way to proceed here is to rule out all possibilities other than lim n →∞ a n = 0.

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