Why is $\lim_{x\to\infty} \frac1x {x!}^{1/x} = \frac1e$?

Question

I've tried various methods, but I couldn't get it equal to $\frac1e$. Please help!

Answer 1

$$A=\lim_{n\to\infty}\dfrac1n (n!)^{1/n}=\lim_{n\to\infty}\left(\prod_{r=1}^n\dfrac rn\right)^{1/n}$$

$$\ln A=\lim_{n\to\infty} \dfrac1n\sum_{r=1}^n\ln\dfrac rn$$

Like The limit of a sum $\sum_{k=1}^n \frac{n}{n^2+k^2}$ use

$$ \lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$

See : integral of $\ln x$ from 0 to 1