$\lim_{ n\to \infty} \frac{n!}{n^n}$ via L'Hospital's rule I just need to find this limit and I don't know how to use L'Hopital's rule in this case:
$$\lim_{ n\to \infty}   \frac{n!}{n^n}.$$
I apologize for the lack of formatting, I've never used the site before.
 A: hint: $\dfrac{n!}{n^n} < \dfrac{1}{n}$
A: Consider the series $\sum\limits_{n=1}^{\infty}\frac{n!}{n^n}$ and set  $a_n=\frac{n!}{n^n}$. By ratio test  $\lim\limits_{n->\infty}\frac{a_{n+1}}{a_n}=\frac{1}{e}<1$. Hence this series is convergent. It follows that $\lim\limits_{n->\infty}\frac{n!}{n^n}=0$.
A: You can use that $n!$'s order of magnitude is lower than $n^n$'s, so if you are watching $\frac{n!}{n^n}$, the limes must be $0$.
The same way you can prove for example, that $\frac {logn}{n}$ has a limit of $0$ too.
Here are the order of magnitude, in order:
$log n << \sqrt n << n << n^x << x^n << n! << n^n$
All of these are going towars $\infty$, but you can see that left to the right they are faster and faster.
If you divide a smaller one with a bigger one, the limit is $0$, and if you divide a bigger one with a smaller one, the limit is $\infty$
Hope this helps a bit. :)
A: Consider the Gamma Function
$$\Gamma(n+1)=\int_0^{\infty}x^ne^{-x}dx$$ For integer values of $n$ we have $\Gamma(n+1)=n!$  
Also for large $n$ values, we can use The Stirling's Approximation
$$\Gamma(n+1)\sim\sqrt{2\pi n}\left(\dfrac{n}{e}\right)^n$$ Therefore
$$\dfrac{\Gamma(n+1)}{n^n}\sim\sqrt{2\pi n}\left(\dfrac{1}{e}\right)^n$$ Using L'Hospital's Rule we can show that 
$$\lim_{n\to\infty}\dfrac{\sqrt n}{e^n}=0$$ Hence
$$\lim_{n\to\infty}\dfrac{n!}{n^n}=0.$$
