HW#4 - SOLUTIONS TO HOMEWORK 4(−1)n = 0 Let n(−1)n 1 =...

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Unformatted text preview: SOLUTIONS TO HOMEWORK # 4 (−1)n = 0. Let n (−1)n 1 = . Thus lim = 0. N n 8.1 (a) lim (b) lim Thus lim 1 n1/3 1 n1/3 > 0. Take N = 1 = 0. Let > 0. Take N = 3 1 . Then n > N ⇒ . Then n > N ⇒ 1 n1/3 (−1)n −0 < n −0 < 1 =. N 1/3 = 0. 2n − 1 2 72 2n − 1 2 = . Let > 0. Take N = − . Then n > N ⇒ −≤ 3n + 2 3 9 3 3n + 2 3 −7 2n − 1 2 < , so lim =. 9n + 6 3n + 2 3 (c) lim n+6 n2 = 0. Let > 0. For n > 3, n + 6 ≤ 7n, n2 − 6 ≥ . So let n2 − 6 2 14 n+6 n+6 14 7n N = max{3, }. Then n > N ⇒ 2 −0 ≤ 1 2 = < , so lim 2 = 0. n −6 n n −6 n 2 (d) lim n 1 n = 0. Let > 0. Take N = . Then n > N ⇒ 2 −0 < +1 n +1 n 1 n = < , so lim 2 = 0. n2 n n +1 8.2 (a) lim n2 7 106 7n − 19 7 7n − 19 = . Let > 0. Let N = . Then n > N ⇒ − ≤ 3n + 7 3 9 3n + 7 3 −106 106 106 7n − 19 7 = < < , so lim =. 9n + 21 9n + 21 9n 3n + 7 3 (b) lim 4n + 3 4 4 41 5 4n + 3 = . Let > 0. Take N = + . Then n > N ⇒ −= 7n − 5 7 49 7 7n − 5 7 41 41 4n + 3 4 = < , so lim =. 49n − 35 49n − 35 7n − 5 7 (c) lim 1 2n + 4 2 16 2 2n + 4 2 = . Let > 0. Take N = − . Then n > N ⇒ −= 5n + 2 5 25 5 5n + 2 5 2n + 4 2 16 16 < , so lim =. = 25n + 10 25n + 10 5n + 2 5 (d) lim (e) lim 1 sin n sin n 1 sin n ≤ < , so lim = 0. Let > 0. Take N = . Then = 0. n n n n 8.4 Let lim sn = 0, |tn | ≤ M for all n. Let that n > N ⇒ |sn | < deﬁnition lim sn tn = 0. M > 0. Since lim sn = 0, ∃ N such . Then n > N ⇒ |sn tn | ≤ |sn ||tn | < 2 M · M = , thus by ...
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HW#4 - SOLUTIONS TO HOMEWORK 4(−1)n = 0 Let n(−1)n 1 =...

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