求limx→0 (e^x-1)sinx/1-cosx
问题描述：
求limx→0 (e^x-1)sinx/1-cosx
答
答：
lim(x→0) (e^x-1)sinx / (1-cosx)
=lim(x→0) xsinx / [2sin ^2 (x/2)]
=lim(x→0) (x^2) / [2*(x/2)^2]
=2
答
limx→0 (e^x-1)sinx/1-cosx
=limx→0[ (e^x))sinx+ (e^x-1)cosx]/sinx
=limx->0e^x+limx-0(e^x-1)cosx]/sinx
=1+limx->0[(e^x)cosx]-(e^x-1)sinx]/cosx
=1+1-0
=2