>
数学
>
求极限limx→0 ∫(0→2x) ln(1+t)dt/x^2
人气:474 ℃ 时间:2020-04-06 10:49:59
解答
limx→0 ∫(0→2x) ln(1+t)dt/x^2
洛必达法则
=lim[x→0] 2ln(1+2x)/(2x)
=lim[x→0] ln(1+2x)/x
等价无穷小代换
=lim[x→0] 2x/x
=2
希望可以帮到你,如果解决了问题,请点下面的"选为满意回答"按钮,
推荐
求极限limx下面是x-->0 ln(1+x)/x
求limx-》0 ∫ln(1+t^2)dt/x^3 积分上限x 下限0
limx→π/2(ln(sinx))/(π-2x)^2求极限
已知limx→0,∫(上限x下限0)(2x-t)ln(1+t)dt/x^n=k,求n
极限limx→0 x/ln(1+x^2)=()
两数和为3,积为-10 求这两个数
用英语简单描写你的房间
以我的卧室写一篇英语作文
猜你喜欢
carefully
After paying 1000 dollars ____,you will all become full memebers of our club A each B all c every
Our class has a map of the world (改为同义句)
The baby didn't cry any more when she saw her mother.同义句
How do you go to school,on foot or by bus?
时间旅行的悖论
设f(x)=ax7+bx5+cx3+dx+5,其中a,b,c,d为常数.若f(-7)=-7,则f(7)=_.
计算1-3+5-7+9-11+…+97-99.
© 2024 79432.Com All Rights Reserved.
电脑版
|
手机版