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