原式=lim(x→0)[e^x-1-ln(1+x)]/[(e^x-1)*ln(1+x)]=lim(x→0)[e^x-1-ln(1+x)]/x²=lim(x→0)[e^x-1/(1+x)]/2x=lim(x→0)[e^x+1/(1+x)²]/2=(1+1)/2=1
