Question

(e^x-1)/x,当x趋于0的极限怎么求啊?

Answer 1

设e^x-1=t 则e^x=1+t x=ln(1+t)
lim[x-->0](e^x-1)/x
=lim[t-->0]t/ln(1+t)
=lim[t-->0]1/ln(1+t)^(1/t)
=1