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
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
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
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