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math131Ahw03 - Homework#3 Spring 2010 Homework solutions by...

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Homework #3 Math 131A Due: Mar. 4 Spring 2010 Prof. Maruskin Homework solutions by: Liem Tran 9.1 Using the limit theorems 9.2-9.6 and 9.7, prove the following. Justify all steps. (a) lim n + 1 n = 1 lim n + 1 n = lim 1 + 1 n = lim(1) + lim 1 n by theorem 9 . 3 = 1 + 0 by 9 . 7 a = 1 (b) lim 3 n + 7 6 n - 5 = 1 2 lim 3 n + 7 6 n - 5 = lim 3 + 7 n 6 - 5 n lim 3 + 7 n = lim(3) + lim 7 n by theorem 9 . 3 = 3 + 0 by 9 . 7 a = 3 lim 6 - 5 n = lim(6) - lim 5 n by theorem 9 . 3 = 6 - 0 by 9 . 7 a = 6 lim 3 n + 7 6 n - 5 = 3 6 by theorem 9 . 6 = 1 2 1
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(c) lim 17 n 5 + 73 n 4 - 18 n 2 + 3 23 n 5 + 13 n 3 = 17 23 lim 17 n 5 + 73 n 4 - 18 n 2 + 3 23 n 5 + 13 n 3 = lim 17 + 73 n - 18 n 3 + 3 n 5 23 + 13 n 3 lim 17 + 73 n - 18 n 3 + 3 n 5 = lim 17 + lim 73 n - lim 18 n 3 + lim 3 n 5 by theorem 9 . 3 = 17 + 73 lim 1 n - 18 lim 1 n 3 + 31 lim 1 n 5 by theorem 9 . 2 = 17 + 0 - 0 + 0 by 9 . 7 a = 17 lim 23 + 13 n 3 = lim 23 + lim 13 n 3 by theorem 9 . 3 = 23 + 13 lim 1 n 3 by theorem 9 . 2 = 23 + 0 by 9 . 7 a = 23 lim 17 n 5 + 73 n 4 - 18 n 2 + 3 23 n 5 + 13 n 3 = 17 23 by theorem 9 . 6 9.9 Suppose that there exists N 0 such that s n t n for all n > N 0 . (a) Prove that if lim s n = + , then lim t n = + . Let M > 0. Since lim s n = + , there exists N N 0 such that s n > M for n > N . Since s n t n then t n > M for all n > N , hence lim t n = + .
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