x^m-1=[(x-1)+1]^m-1
因为x-1趋于0,所以利用等价无穷小〖(x+1)〗^n-1 nx 可得：
[(x-1)+1]^m-1 与 m(x-1)等价
同理可得(x^n-1)等价于n(x-1)
lim(x^m-1)/(x^n-1)
=lim[m(x-1)]/[n(x-1)]
=m/n