求Lim(x->0)(sinx-e^x+1)/[1-(1-x^2)^1/2] 的极限

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求Lim(x->0)(sinx-e^x+1)/[1-(1-x^2)^1/2] 的极限
求Lim(x->0)(sinx-e^x+1)/[1-(1-x^2)^1/2] 的极限

求Lim(x->0)(sinx-e^x+1)/[1-(1-x^2)^1/2] 的极限
J=(sinx - e^x +1)/[1-(1-x^2)^0.5]
用洛必达法则:
lim(x->0)J=lim(x->0) (cosx - e^x)/[x/(1-x^2)^0.5]
=lim(x->0) (cosx - e^x)(1-x^2)^0.5/x //:还得用一次洛必达法则
=lim(x->0) [(-sinx - e^x)(1-x^2)^0.5+(cosx - e^x)*0.5*(-2x)/(1-x^2)^0.5]
=-1+0
=-1

Lim(x->0)(sinx-e^x+1)/[1-(1-x^2)^1/2]
=Lim(x->0)(cosx-e^x)/[x/√(1-x^2)]=Lim(x->0)(cosx-e^x)/x
=Lim(x->0)(-sinx-e^x)
=0-1=-1为什么[1-(1-x^2)^1/2]求导得出[x/√(1-x^2)]??[1-(1-x^2)^1/2]'=-(1/2)(1-x^2...

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Lim(x->0)(sinx-e^x+1)/[1-(1-x^2)^1/2]
=Lim(x->0)(cosx-e^x)/[x/√(1-x^2)]=Lim(x->0)(cosx-e^x)/x
=Lim(x->0)(-sinx-e^x)
=0-1=-1

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