![](https://iknow-base.cdn.bcebos.com/lxb/notice.png)
展开全部
(8)
lim(x->0) ln(1+ 1/x)/arccotx (0/0)
分子,分族祥母分别求导
=lim(x->0) [x/(x+1)](-1/x^2)/ [ - 1/(1+x^2) ]
=lim(x->陪唯0) (1+x^2)/[ x(x+1) ]
=lim(x->0) (1/x^2 +1)/(1+1/x)
=1
(9)
x->0
sinx~ x - (1/6)x^3
sin2x ~ 2x - (4/3)x^3
(sin2x)^2
~ [2x - (4/3)x^3]^2
~ 4x^2 - (16/3)x^4
x^2 - (1/4)(sin2x)^2 ~ (4/兆乱搏3)x^4
lim(x->0) [ 1/(sinx)^2 - (cosx)^2/x^2 ]
=lim(x->0) [x^2 - (sinx.cosx)^2 ]/[x^2.(sinx)^2]
=lim(x->0) [x^2 - (1/4)(sin2x)^2 ]/[x^2.(sinx)^2]
=lim(x->0) (3/4)x^4/x^4
=3/4
lim(x->0) ln(1+ 1/x)/arccotx (0/0)
分子,分族祥母分别求导
=lim(x->0) [x/(x+1)](-1/x^2)/ [ - 1/(1+x^2) ]
=lim(x->陪唯0) (1+x^2)/[ x(x+1) ]
=lim(x->0) (1/x^2 +1)/(1+1/x)
=1
(9)
x->0
sinx~ x - (1/6)x^3
sin2x ~ 2x - (4/3)x^3
(sin2x)^2
~ [2x - (4/3)x^3]^2
~ 4x^2 - (16/3)x^4
x^2 - (1/4)(sin2x)^2 ~ (4/兆乱搏3)x^4
lim(x->0) [ 1/(sinx)^2 - (cosx)^2/x^2 ]
=lim(x->0) [x^2 - (sinx.cosx)^2 ]/[x^2.(sinx)^2]
=lim(x->0) [x^2 - (1/4)(sin2x)^2 ]/[x^2.(sinx)^2]
=lim(x->0) (3/4)x^4/x^4
=3/4
追问
第8题不是无穷比无穷吗,为啥等于1
追答
(8)
lim(x->0) ln(1+ 1/x)/arccotx (0/0)
=lim(x->0) d/dx .ln(1+ 1/x)/ d/dx .arccotx
分母->0, 分子->0
分子,分母分别求导, 洛必达
=lim(x->0) [x/(x+1)](-1/x^2)/ [ - 1/(1+x^2) ]
=lim(x->0) (1+x^2)/[ x(x+1) ]
=lim(x->0) (1/x^2 +1)/(1+1/x)
=1
本回答被网友采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询