求各極限值 問題
1.lim ln(e^x-1)/lnx
x→0^+
2.lim (1/sin^2x -1/x^2)
x→0
3.lim (cot)^1/lnx
x→0^+
4.lim (x+e^x)^1/x
x→0
5.lim x^2(1-cos4/x)
x→○○
微積分高手幫忙一下
2013-08-25 9:39 am
回答 (2)
2013-08-25 6:00 pm
✔ 最佳答案
﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣圖片參考：http://imgcld.yimg.com/8/n/HA05107138/o/20130825095914.jpg
圖片參考：http://imgcld.yimg.com/8/n/HA05107138/o/20130825095946.jpg
2013-08-28 10:42 pm
只供參考
解法三:
Lim [1/(Sinx)^2 – 1/x^2]
=Lim [x^2-(Sinx)^2]/[x^2*(Sinx)^2]
=Lim [(x+Sinx)(x-Sinx)]/[x^4*(Sinx/x)^2]
=Lim [(x+Sinx)/x]*[(x-Sinx)/x^3]*[1/ (Sinx/x)^2]
{由洛必達法則:
Lim [(x+Sinx)/x]=2
Lim [(x-Sinx)/x^3]=1/6
Lim [1/ (Sinx/x)^2]=1/[Lim (Sinx/x)]^2=1 }
=2*1/6*1=1/3
解法三:
Lim [1/(Sinx)^2 – 1/x^2]
=Lim [x^2-(Sinx)^2]/[x^2*(Sinx)^2]
=Lim [(x+Sinx)(x-Sinx)]/[x^4*(Sinx/x)^2]
=Lim [(x+Sinx)/x]*[(x-Sinx)/x^3]*[1/ (Sinx/x)^2]
{由洛必達法則:
Lim [(x+Sinx)/x]=2
Lim [(x-Sinx)/x^3]=1/6
Lim [1/ (Sinx/x)^2]=1/[Lim (Sinx/x)]^2=1 }
=2*1/6*1=1/3
收錄日期: 2021-04-23 23:27:34
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130825000015KK00574