How to find $\lim\limits_{n\to\infty}\frac{n^2}{n!}$ 
Find the limit of $\lim\limits_{n\to\infty}\frac{n^2}{n!}$.

There is a way to note that if $n>4$ then
$$0 < \frac{n^2}{n!} < \frac{n}{(n-2)(n-1)} < \frac{1}{n-3}$$
 and consequently the limit is zero. But can somebody provide another way to evaluate this limit?
 A: If $n\geq2$ then you can evaluate it this way:
\begin{align*}
\lim_{n\to\infty} \frac{n^2}{n!}&=\lim_{n\to\infty} \frac{n}{(n-1)!}\\
&=\lim_{n\to\infty} \frac{n-1+1}{(n-1)!}\\
&=\lim_{n\to\infty} \bigg(\frac{n-1}{(n-1)!}+\frac{1}{(n-1)!}\bigg)\\
&=\lim_{n\to\infty} \bigg(\frac{1}{(n-2)!}+\frac{1}{(n-1)!}\bigg)\\
&=\lim_{n\to\infty} \frac{1}{(n-2)!}+\lim_{n\to\infty} \frac{1}{(n-1)!}\\
&=0
\end{align*}
A: Hint: This limit is closely related to derivatives of the power series for the exponential function. 
Solution:

 $\displaystyle\sum_{n=1}^\infty \frac{n^2 z^n}{n!} = e^z (z^2+z)$ and so $\displaystyle\sum_{n=1}^\infty \frac{n^2}{n!} = 2e$. Since the series converges, we must have $\frac{n^2}{n!} \to 0$.

A: If you know that $n^2<2^n$ for $n>5$ you can look at $\sum\limits_{n=1}^{\infty} \frac{n^2}{n!}$. As $\sum\limits_{n=0}^\infty \frac{2^n}{n!}=e^2$ by the comparison test the series $\sum\limits_{n=1}^{\infty} \frac{n^2}{n!}$ must converge and therefore $\lim\limits_{n\to\infty} \frac{n^2}{n!}=0$.
