$\displaystyle\lim_{n\rightarrow\infty}\left\{\dfrac{n!}{(kn)^n}\right\}^{\dfrac{1}{n}}, k\neq 0$, is equal to?

$\begin{array}{l} \lim _{n \rightarrow \infty}\left\{\frac{n !}{(k n)^{n}}\right\}^{1 / n} \\ L=\lim _{n \rightarrow \infty}\left\{\frac{n !}{(k n)^{n}}\right\}^{1 / n} \\ \log L=\lim _{n \rightarrow \infty} \frac{1}{n} \ln \left(\frac{n !}{(k n)^{n}}\right) \\ =\lim _{n \rightarrow \infty} \frac{\ln \left(\frac{n}{k n} \times \frac{n-1}{k n} \times \ldots \times \frac{2}{k n} \times \frac{1}{k n}\right)}{n} \\ =\lim _{n \rightarrow \infty} \frac{\left[\ln \left(\frac{n}{k n}\right)+\ln \left(\frac{n-1}{k n}\right)+\cdots \log \left(\frac{1}{k n}\right)\right]}{n} \\ \frac{1}{n}=d x \quad \frac{r}{n}=\frac{n}{n} \\ \end{array}$

Take $\log$ on both sides

$\log L=\lim _{n \rightarrow \infty} \frac{1}{n} \ln \left(\frac{n !}{(k n)^{n}}\right)$

Using Integration as the limit of sum:

$\begin{array}{l} =\lim _{n \rightarrow \infty} \sum_{r=1}^{n}\left(\frac{1}{n}\right) \log \left(\frac{r}{k n}\right) \\ =\int_{0}^{1} \ln \left(\frac{x}{k}\right) d x \text {. [using by parts] } \\ =\left.x \ln \left(\frac{x}{k}\right)\right|_{0} ^{1}-\int_{0}^{1} \frac{1}{x} x d x \\ =\ln \left(\frac{1}{k}\right)-0-1=-\ln k-1 \\ =-\ln (k e) \\ \log L=-\log (k e) \\ \log 12 \log \left(\frac{1}{k l}\right) \\ L=\frac{1}{k e} \text { } \\ \end{array}$