大一數學，關於極限(limit)問題...唔該幫幫忙>

小星S

2009-09-29 7:11 am

求limit...

(1). lim x->∞ ( x + 1 / x - 1 ) ^x+1

(2). lim x->∞ ( x^2 / x^2 - 1 ) ^x

有一條公式的 : lim x->∞ ( 1 + 1 / x ) ^x = e
<=> lim x->0 ( 1 + x )^1/x = e

唔該幫幫忙><~~我知第1題ans係 e^2....第2題ans係 1...
但我解左好耐都唔明點解出呢個ans出黎><...

識ge話麻煩解答下=3=''....thxx!!!

回答 (1)

用戶9112

2009-09-29 7:42 am

✔ 最佳答案

lim x->∞ [(x + 1) / (x - 1)]^(x + 1)
= lim x->∞ [1 + 2 / (x - 1)]^(x - 1 + 2)
= lim x->∞ [1 + 2 / (x - 1)]^(x - 1)[1 + 2 / (x - 1)]^2
= {lim x->∞ [1 + 2 / (x - 1)]^(x - 1)}{lim x->infinity[1 + 2 / (x - 1)]^2}
= (e^2)(1)
= e^2
lim x->∞ [x^2 / (x^2 - 1)]^x

= lim x->∞ [x^2 / (x^2 - 1)]^(x^2 - 1 + 1)/x

= lim x->∞ [1 + 1 / (x^2 - 1)]^(x^2 - 1)/x[1 + 1 / (x^2 - 1)]^(1/x)
= {lim x->∞ [1 + 1 / (x^2 - 1)]^(x^2 - 1)/x}{lim x->∞[1 + 1 / (x^2 - 1)]^(1/x)}
= lim x->∞ {[1 + 1 / (x^2 - 1)]^(x^2 - 1)}^(1/x){1^0}
= lim x->∞ {e}^(1/x)
= e^0
= 1

👍 0 👎 0