题目
用洛必塔法则 求极限lim x趋于0 e^(sinx)-e^x/sinx-x 这个极限为什么等于1呢?
提问时间：2021-01-02
答案
lim[e^(sinx)-e^x]/(sinx-x)=lim [e^(sinx)*cosx-e^x]/(cosx-1)=
x->0 x->0
lim [e^(sinx)*(cosx)^2-e^(sinx)*sinx-e^x]/(-sinx)=
x->0
lim [e^(sinx)*(cosx)^3-e^(sinx)*2cosxsinx-e^(sinx)*sinxcosx-e^(sinx)*cosx-e^x]/(-cosx)=
x->0
=(1-0-0-1-1)/(-1)=1
x->0 x->0
lim [e^(sinx)*(cosx)^2-e^(sinx)*sinx-e^x]/(-sinx)=
x->0
lim [e^(sinx)*(cosx)^3-e^(sinx)*2cosxsinx-e^(sinx)*sinxcosx-e^(sinx)*cosx-e^x]/(-cosx)=
x->0
=(1-0-0-1-1)/(-1)=1
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