Prove that lim n! n n = 0

$\underset{}{\lim f\left(n\right)}=\underset{}{\lim }\left(\frac{n!}{n^n}\right)$

We know that $h\left(\underset{}{\lim } g\left(x\right)\right)=\underset{}{\lim }f\left(g\left(x\right)\right)$

Therefore

$$\begin{gathered} \log \left(\underset{}{\lim }f\left(n\right)\right)=\underset{}{\lim }\log \left(\frac{n!}{n^n}\right) \\ =\underset{}{\lim }\log \left(\frac{1}{n}\right)+\underset{}{\lim }\log \left(\frac{2}{n}\right)+.....+\underset{}{\lim }\log \left(\frac{n}{n}\right) \end{gathered}$$

Now we have to show that  for $1\le m\le n$, $\underset{}{\lim }\log \left(\frac{m}{n}\right)=0$ holds which is enough to say that $\underset{}{\lim f\left(n\right)}=\underset{}{\lim }\left(\frac{n!}{n^n}\right)=0$

From the definition of of limit  that a function $f$ is said to be  tends to $l$ as $x\to \infty$ if for each $\epsilon >0$ there exist $k>0$ such that $\left|f\left(x\right)-l\right|<\epsilon ,$ when $x>k$

Let choose a positive $\epsilon$

Now $\left|\log \left(\frac{m}{n}\right)-0\right|<\epsilon$ will hold if $\frac{m}{n}<e^{\epsilon },$ that is $n>\frac{m}{e^{\epsilon }}$ .

Now let $k=\frac{m}{e^{\epsilon }}$ and we know $1\le m\le n$ and $e^{\epsilon }>1$, clearly $\frac{m}{e^k}<n$, therefore $n>k$

Therefore for  $\left|\log \left(\frac{m}{n}\right)-0\right|<\epsilon$ there exist $n\ge k$

Hence the limit exist and the function $f\left(n\right)\to 0$ as $n\to \infty$ if for each $\epsilon >0$ there exist $k>0$ such that $\left|f\left(n\right)-0\right|<\epsilon ,$ when $n>k$

$\therefore \underset{n\to \infty }{\lim }\log \left(\frac{m}{n}\right)=0$

Therefore $\underset{}{\lim f\left(n\right)}=\underset{}{\lim }\left(\frac{n!}{n^n}\right)=0$