$\lim _{x \rightarrow \infty} \frac{(n !)^{1 / n}}{n}=\lim _{n \rightarrow-}\left(\frac{n !}{n^{n}}\right)^{1 / n}$
We have, $\frac{n !}{n^{n}}=\frac{1 \cdot 2 \cdot 3}{n \cdot n \cdot n} \frac{n}{n}$
$\therefore \quad\left\{\frac{n !}{n^{n}}\right\}^{1 / n}=\left\{\frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdots \frac{r}{n} \cdots \frac{n}{n}\right\}^{1 / n}$
$\Rightarrow \lim _{n \rightarrow \infty}\left\{\frac{n !}{n^{n}}\right\}^{1 / n}=\lim _{n \rightarrow \infty}\left\{\frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdots \frac{r}{n} \cdots \frac{n}{n}\right\}^{1 / n}$
Let $\quad A=\lim _{n \rightarrow \infty}\left\{\frac{n !}{n^{n}}\right\}^{1 / n}$
Then, $A=\lim _{n \rightarrow-}\left\{\frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdots \frac{r}{n} \cdots \frac{n}{n}\right\}^{1 / n}$
$\begin{aligned} \Rightarrow \quad \log A &=\lim _{n \rightarrow \infty} \frac{1}{n} \Sigma \log \left(\frac{r}{n}\right)=\int_{0}^{1} \log x d x \\ &=\left[x \log x-\int \frac{1}{x} \cdot x d x\right]_{0}^{1} \end{aligned}$
Integrating by parts $=[x \log x-x]_{0}^{1}=-1$
$\Rightarrow \quad A=e^{-1}=\frac{1}{e}$