Possible Duplicate:
Showing that $\frac{\sqrt[n]{n!}}{n}$ $\rightarrow \frac{1}{e}$
I have a litle discution with a friend about the folowing limit: $\lim_{n\to\infty} \frac{\sqrt[n]{n!}}{n}$ I would solve it like this: $\lim_{n\to\infty} \sqrt[n]{\frac{n!}{n^{n}}} =0$ or $\lim_{n\to\infty} \frac{\sqrt[n]{n}\sqrt[n]{n-1}\cdots\sqrt[n]{1}}{n}=\frac{1*1*1\cdots}{\infty}=0$ and in this 2º way would there be a problem with $1^{\infty}$? I would say that no, because there is no functions involved, since as much as I know this undetermination is because you would whant to avoid the situation such as $f(x)^{g(x)}$ where $f(x)\to1$ and $g(x)\to\infty$ Could anyone clarify this for me?