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Unformatted text preview: Solutions for Test 2 1 . State the definition that a sequence ( s n ) converges to a number L . For any > 0 there is N such that  s n L  < whenever n > N . 2 . Find the limit of each sequence if it converges: (a) x n = n 2 3 n lim x n = 0 (b) y n = 3 n 2 7 n 2 + 5 lim y n = 3 / 7 (c) z n = 2 n 1 lim z n does not exist (lim z n = ∞ ) 3 . Prove that the sequence ( s n ), where s n = 1 √ n , converges to 0. Let > 0. Let N = 1 / 2 . Then if n > N we have n > 1 / 2 so 1 / √ n < . Therefore,  s n  = 1 / √ n < whenever n > N so we are done. 4 . Prove that the sequence ( x n ), where x n = n 2 n 2 1 , converges to 0. Note that n 2 n 2 1 < 1 n if n > 1. [Show this as follows: 1 < n implies 1 < n 2 so n 2 < 2 n 2 1. So n 2 n 2 1 < 1 n . Note these steps are the reverse of simplying the inequality itself.] So here is the proof itself: Let > 0. Let N = max { 1 , 1 / } . Then if n > N , we have n > 1 / so 1 /n < . Therefore for n > N we have n 2 n 2 1 0 = n 2 n 2 1 < 1 n < and...
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This note was uploaded on 08/24/2009 for the course MATH 262447221 taught by Professor Weisbart during the Spring '09 term at UCLA.
 Spring '09
 WEISBART

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