求limx→0 (e^x-1)sinx/1-cosx

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求limx→0 (e^x-1)sinx/1-cosx
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求limx→0 (e^x-1)sinx/1-cosx
求limx→0 (e^x-1)sinx/1-cosx

求limx→0 (e^x-1)sinx/1-cosx
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

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