Question

limx趋进0,求极限e^x-1/x,令t=e^x-1

Answer 1

令t=e^x-1,x=ln(t+1)
原式=t/ln(t+1)
=1/[(1/t)ln(t+1)]
=1/ln(1+t)^(1/t)(t->0)
=1/lne
=1
解法2
原式=(e^x-1)/x(x->0)=（e^x-1)'/x'(罗比塔法则）=e^x/1=e^0=1
殊途同归