当x趋向于正无穷时,1/x-1与1/x^2-1比较两个无穷小的阶求步骤
问题描述:
当x趋向于正无穷时,1/x-1与1/x^2-1比较两个无穷小的阶求步骤
答
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