![](https://iknow-base.cdn.bcebos.com/lxb/notice.png)
lim n->无穷 (n+1)^(1/n)=
展开全部
设f(x)=(x+1)^(1/x)
∵lim(x->∞)f(x)=lim(x->∞)[(x+1)^(1/x)]
=lim(x->∞){e^[ln(x+1)/x]}
=e^{lim(x->∞)[ln(x+1)/x]}
=e^{lim(x->∞)[1/(x+1)] (0/0型,应用罗比达法则)
=e^0
=1
∴对于自然数n,也一定有
lim(n->∞)[(n+1)^(1/n)]=1,成立.
∵lim(x->∞)f(x)=lim(x->∞)[(x+1)^(1/x)]
=lim(x->∞){e^[ln(x+1)/x]}
=e^{lim(x->∞)[ln(x+1)/x]}
=e^{lim(x->∞)[1/(x+1)] (0/0型,应用罗比达法则)
=e^0
=1
∴对于自然数n,也一定有
lim(n->∞)[(n+1)^(1/n)]=1,成立.
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询