$\underset{n\to \infty }{Lt} {\left(\frac{{\displaystyle n!}}{{\displaystyle n^n}}\right)}^{\frac{1}{n}}=$

$Let y=\underset{n\to \infty }{Lt} {\left(\frac{{\displaystyle n!}}{{\displaystyle n^n}}\right)}^{\frac{1}{n}}$

$$\begin{gathered} \log y=\underset{n\to \infty }{Lt} \log {\left(\frac{n!}{n^n}\right)}^{1/n} \\ =\underset{n\to \infty }{Lt}\frac{1}{n}\log \left(\frac{1.2.3......n}{n.n.n.....n}\right) \\ \log y=\underset{n\to \infty }{Lt} \log {\left(\frac{n}{n^n}\right)}^{1/n} \\ =\underset{n\to \infty }{Lt}\frac{1}{n}\log \left(\log \frac{1}{n}+\log \frac{2}{n}+....+\log \frac{n}{n}\right) \\ =\underset{n\to \infty }{Lt}\frac{1}{n}\sum _{n=1}^n\log \frac{r}{n} \\ \begin{array}{l}=\underset{0}{\overset{1}{\int }}\log xdx\\ ={(x\log x-x)}_0^1\\ =(\log 1-1)-0\\ =-1\end{array} \end{gathered}$$

$\left[\begin{array}{l}\therefore x\to 0\Rightarrow x\log x\to 0\\ because\underset{x\to 0}{Lt}x\log x\\ =\underset{x\to 0}{Lt}\frac{{\displaystyle \log x}}{{\displaystyle 1/x}}\frac{\infty }{\infty }form\\ =\underset{x\to 0}{Lt}(-x)=0\end{array}\right]$