**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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- Fall '08
- SURGENT
- Calculus, Squeeze Theorem, Limit, lim