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Unformatted text preview: Math 131A  Section 2 Spring 2010 Homework 3 Problems graded: 8.6 (2 points), 8.8 (3 points), 8.10 (2 points), 9.3 (2 points), 9.6 (3 points), 9.11 (3 points) 8.2(a). Let > 0 and set N = 1 . Then, n n 2 + 1 0 = n n 2 + 1 ≤ n n 2 = 1 n < for n > N . Hence, lim n →∞ n n 2 +1 = 0. (c). Let > 0 and set N = max { 1 , 42 41 1 } . Then, 4 n + 1 7 n 5 4 7 = 41 7(7 n 5) , and for n > N ≥ 5, we have that 7 n 5 > 6 n , and so 41 7(7 n 5) < 41 42 n < . We conclude that lim n →∞ 4 n +3 7 n 5 = 4 7 . (e). Let > 0 and set N = 1 . Then, for n > N , we have that 1 n sin( n ) 0 = 1 n  sin( n )  ≤ 1 n < since  sin( x )  ≤ 1 for all x ∈ R . Therefore, lim n →∞ 1 n sin( n ) = 0. 8.3. Let > 0. Since lim n →∞ s n = 0, there is N such that  s n  < 2 for all n > N . Then, for n > N , we have that  √ s n  = √ s n < , so lim n →∞ √ s n = 0. 8.6.(a). Suppose lim n →∞ s n = 0, and let > 0. Choose N such that  s n  < for n > N . Then,  s n    =  s n  =  s n  < for n > N , so lim n →∞  s n  = 0. Now suppose lim n →∞  s n  = 0 and let > 0. Choose N such that  s n   < for n > N . Then,  s n ...
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This note was uploaded on 11/20/2010 for the course MATH 131A 131A taught by Professor Kim during the Spring '10 term at UCLA.
 Spring '10
 Kim

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