x-> 0+
更新1:
點解 [1/ln(10)] lim [x->0+] ln(x) / (1/x) = [1/ln(10)] lim [x->0+] (1/x) / -(1/x^2)
急! 1題 limit, 要show work
回答 (1)
2009-10-09 11:51 pm
✔ 最佳答案
log(x) = ln(x) / ln(10)
lim [x->0+] xlog(x)
= lim [x->0+] xln(x) / ln(10)
= [1/ln(10)] lim [x->0+] ln(x) / (1/x)
= [1/ln(10)] lim [x->0+] (1/x) / -(1/x^2)
= [1/ln(10)] lim [x->0+] (-x)
= 0
2009-10-09 16:49:57 補充：
可參考http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule
l'Hôpital's rule
收錄日期: 2021-04-24 10:15:29
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091009000051KK00648