∵lim(x->0)[ln(x+e^x)/x]=lim(x->0)[(1+e^x)/(x+e^x)] (0/0型極限,應用羅比達法則)
=(1+1)/(0+1)
=2
∴l(xiāng)im(x->0)[(x+e^x)^(2/x)]=lim(x->0){e^[(2/x)ln(x+e^x)]}
=e^{2*lim(x->0)[ln(x+e^x)/x]}
=e^(2*2)
=e^4.