lim(x趋向于0)(e^x2+cosx-1)^(1/x2)
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lim(x趋向于0)(e^x2+cosx-1)^(1/x2)
lim(x趋向于0)(e^x2+cosx-1)^(1/x2)
lim(x趋向于0)(e^x2+cosx-1)^(1/x2)
这是一个1^无穷的极限
(e^x2+cosx-1)^(1/x2)=e^(ln(e^x2+cosx-1))/x2
所以lim(x→0)(e^x2+cosx-1)^(1/x2)
=lim(x→0)e^(ln(e^x2+cosx-1))/x2
=e^lim(x→0)(ln(e^x2+cosx-1))/x2
=e^lim(x→0)(xe^x-sinx)/x(e^x2+cosx-1) (洛必达法则)
=e^lim(x→0)(e^x+xe^x-cosx)/((e^x2+cosx-1)+x(2xe^x-sinx))(洛必达法则)
=e^0=1
设y=(e^x2+cosx-1)^(1/x2)
则limlny=lim(ln(e^x2+cosx-1))/x^2
=lim(2xe^x2-sinx)/(2x(e^x2+cosx-1)) [分子分母分别求导]
=lim(2e^x2+2x()-cosx)/(2(e^x2+cosx-1)+2x()) [再次分子分母分别求导]
=1/2,即limy=e^(1/2)
利用两次洛必达法则解决问题
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