thx
lim(x-->1) e^x-ex/ (x-1)^2
ans是1.36
請幫幫忙,感激!
更新1:
唔好意思啊, 我未學L'Hopital's Rule 而微分都係岩岩先接觸,唔係太明~_~
更新2:
回知識長,,學左冇耐^_^ 點chop::等我消化下先,感謝^_^
maths (limit問題)急!!
回答 (2)
2010-10-18 8:07 am
✔ 最佳答案
The given limit is in the form 0/0lim e^x -ex / (x - 1)^2
= lim e^x -e / 2(x - 1) (L'Hopital's Rule)
= lim e^x/2 (L'Hopital's Rule)
= e/2 = 1.359140914 = 1.36
2010-10-18 16:51:16 補充：
or,alternatively
e^x = (e)(e^x-1)
= e[1 + (x - 1) + (x - 1)^2/2! + (x - 1)^3/3!+...]
So lim (e^x - ex)/(x - 1)^2
= lim e[(x - 1)^2/2! + (x - 1)^3/3! + (x - 1)^4/4! + ...]/(x - 1)^2
= lim e/2! + e[(x - 1)/3! + (x - 1)^2/4! + ...]
= e/2
= 1.36
參考: me
2010-10-18 5:46 pm
有無學過e^X=1+x+x^2/2!+x^3/3!+…?
收錄日期: 2021-04-23 23:23:25
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20101017000051KK01596