极限习题问题 为什么(x趋向0)lim(1-cos)/[x-ln(1+x)]=sinx/[1-1/(1+x)]=lim(sinx/x)(1+x)要详细说明

\u6c42\u6781\u9650lim(1-cos(1-cosX))/(sinx^2*ln(1+x^2))\u6709\u56fe\u3002\u3002\u3002\u6c42\u5927\u795e\u5e2e\u5fd9\uff0c\u8c22\u8c22~

\u7b49\u4ef7\u65e0\u7a77\u5c0f\u4ee3\u6362sinx~x,ln(1+x)~x,1-cosx~0.5x^2
\u539f\u5f0f=lim0.5(1-cosx)^2/x^4=lim0.5*(0.5x^2)^2/x^4=1/8

x\u21920:lim ln(sinx/x)=ln lim(sinx/x)=ln1=0

(x→0)lim[(1-cosx)/[x-ln(1+x)]]
是0/0型，故用罗比塔法则，即先对分子、分母分别求导，再求极限
(x→0)lim[(1-cosx)/[x-ln(1+x)]]
=(x→0)lim [sinx/[1-1/(1+x)]]
=(x→0)lim(sinx/x)(1+x)
=1