pure maths

2009-11-21 4:33 am

回答 (1)

2009-11-21 4:54 am
✔ 最佳答案

(a) lim n->∞ (1 - 1/n)^(-n)
= lim n->∞ 1 / [1 + 1/(n-1)]^(-n)
= lim n->∞ [1 + 1/(n-1)]^(n)
= lim n->∞ [1 + 1/(n-1)][1 + 1/(n-1)]^(n - 1)
= lim n->∞ [1 + 1/(n-1)]lim n->∞[1 + 1/(n-1)]^(n - 1)
= (1)(e)
= e
(b)(i) lim n->∞ [1 + 1/(n+1)]^(-n)
= lim n->∞ [1 + 1/(n+1)][1 + 1/(n+1)]^[-(n+1)]
= lim n->∞ [1 + 1/(n+1)]lim n->∞[1 + 1/(n+1)]^[-(n+1)]
= (1)lim n->∞1 / [1 + 1/(n+1)]^(n+1)
= 1/e
(ii) lim n->∞(1 - 4/n – 5/n^2)^n
= lim n->∞[(1 - 5/n)^n][(1 + 1/n)^n]
= lim n->∞[(1 - 5/n)^n]lim n->∞[(1 + 1/n)^n]
= lim n->∞[(1 - 5/n)^n/5]^5(e)
= (1/e)^5(e)
= e^(-4)


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