lim(x~無窮大) (x*log x)/[(x+1)*log(x+1)] =1
要詳解
這題極限（到無窮大）要怎算？（要詳解）
2012-01-22 6:53 am
回答 (2)
2012-01-22 8:50 am
2012-01-22 8:10 am
lim(x->∞) x lnx/[(x + 1)ln(x + 1)]
= lim(x->∞) (x lnx)'/[(x + 1)ln(x + 1)]'
= lim(x->∞) (1 + lnx)/[1 + ln(x + 1)]
= lim(x->∞) (1 + lnx)'/[1 + ln(x + 1)]'
= lim(x->∞) (1/x)/[1/(x + 1)]
= lim(x->∞) (x + 1)/x
= lim(x->∞) 1 + 1/x
= 1
= lim(x->∞) (x lnx)'/[(x + 1)ln(x + 1)]'
= lim(x->∞) (1 + lnx)/[1 + ln(x + 1)]
= lim(x->∞) (1 + lnx)'/[1 + ln(x + 1)]'
= lim(x->∞) (1/x)/[1/(x + 1)]
= lim(x->∞) (x + 1)/x
= lim(x->∞) 1 + 1/x
= 1
收錄日期: 2021-04-26 19:17:47
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120121000015KK07270