271_HW8.2 - MAT 271 Summer 2011 Calculus II Scott Zinzer...

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MAT 271, Summer 2011 Calculus II, Scott Zinzer 8.2 HW 8.2. Sequences. 11. Find the limit of the sequence 3 n 3 - 1 2 n 3 - 1 . Solution: Dividing numerator and denominator by n 3 gives that lim n →∞ 3 n 3 - 1 2 n 3 - 1 = lim n →∞ 3 n 3 /n 3 - 1 /n 3 2 n 3 /n 3 - 1 /n 3 = lim n →∞ 3 - 1 /n 3 2 - 1 /n 3 = 3 - 0 2 - 0 = 3 2 18. Find the limit of the sequence ln(1 /n ) n . Solution: Recall ln(1 /n ) = - ln n . Let f ( x ) = - ln x x . Using L’Hˆopital’s Rule, we obtain lim x →∞ - ln x x = lim x →∞ - 1 x 1 = 0 By Theorem 8.1, lim n →∞ ln(1 /n ) n = 0. 31. Find the limit of the sequence e - n cos n . Solution: We will use the Squeeze Theorem. Since - 1 cos n 1 and e - n is positive, we have - e - n e - n cos n e - n
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Unformatted text preview: lim x →∞-e-x = 0 and lim x →∞ e-x = 0 so by Theorem 8.1 also lim n →∞-e-n = 0 and lim n →∞ e-n = 0 By the Squeeze Theorem, we have lim n →∞ e-n cos n = 0. MAT 271, Summer 2011 Calculus II, Scott Zinzer 8.2 HW 43. Find the limit of ± sin n 2 n ² Solution: We will use the Squeeze Theorem. Now-1 ≤ sin n ≤ 1 so that-1 2 n ≤ sin n 2 n ≤ 1 2 n Now lim n →∞-1 2 n = 0 and lim n →∞ 1 2 n = 0, so by the Squeeze Theorem, lim n →∞ sin n 2 n = 0....
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