前者是比后者低阶的无穷小x→+∞
lim［1/(x-1)］/［1/(x-1)］
=lim(x+1)
=+∞
所以
1/(x-1)是比1/(x-1)低阶的无穷小
或者这样算
x→+∞
lim［1/(x-1)］/［1/(x-1)］
=lim1/(x+1)
=0
所以
1/(x-1)是比1/(x-1)高阶的无穷小