Prove that
lim[n→0] (k^n -1)/(n) = log k
Limit :::::::::::::::::::::::
2012-08-17 8:19 pm
回答 (2)
2012-08-18 12:50 am
✔ 最佳答案
Let t = kⁿ - 1 , then n = logk (1+t) , when n→0 , t→0 , ∴
lim(n→0) (kⁿ - 1) / n
= lim(t→0) t / logk (1+t)
= lim(t→0) 1 / [1/t logk (1+t)]
= lim(t→0) 1 / logk (1+t)¹ˡ ᵗ
= 1 / lim(t→0) logk (1+t)¹ˡ ᵗ
= 1 / logk [lim(t→0) (1+t)¹ˡ ᵗ]
= 1 / logk e
= log k / log e
= ln k
2012-08-17 8:54 pm
Try L'hospital rule.
2012-08-17 16:10:11 補充：
Applying L'hospital rule, differentiate the fraction respect to n.
lim(n→0) (k^n -1)/n = lim(n→0) kⁿ ln k = ln k
2012-08-17 16:10:11 補充：
Applying L'hospital rule, differentiate the fraction respect to n.
lim(n→0) (k^n -1)/n = lim(n→0) kⁿ ln k = ln k
收錄日期: 2021-04-13 18:55:17
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https://hk.answers.yahoo.com/question/index?qid=20120817000051KK00279