lim(x→0)[1/x-1/(e^x-1)]=lim(x→0)[(e^x-1-x)/(x(e^x-1))](这是0/0型,运用洛必达法则））=lim(x→0)[(e^x-1)/(e^x+xe^x-1)]（再运用洛必达法则））=lim(x→0)e^x/(e^x+e^x+xe^x)=1/2
