Find the limit of $\lim_{n\to \infty}\frac{2^n n^2}{n!}$ I am trying to find the limit of the following, $$\lim_{n\to \infty}\frac{2^n n^2}{n!}$$
L'Hospital is not going to work. Hoping for a squeeze, by the observation that $2^n<n!$ for $n\geq 4$ does not help either as one side of the limit goes to $\infty$. How can I solve this? Any hints?

$n!>2\cdot 3^{n-2}$ for $n\ge2$ is apparent from the product definition, so:
$$\frac{2^nn^2}{n!}\le\frac{2^nn^2}{2\cdot 3^{n-2}}=\frac{9n^2}{4\left(\frac{3}{2}\right)^n}$$
Note that $\left(\frac{3}{2}\right)^n=\left(1+\frac{1}{2}\right)^n=\sum_{k=0}^n\binom{n}{k}\left(\frac{1}{2}\right)^k\ge\binom{n}{3}\left(\frac{1}{2}\right)^3$, so we have:
$$\frac{2^nn^2}{n!}\le\frac{9n^2}{4\left(\frac{3}{2}\right)^n}\le\frac{9n^2}{4\binom{n}{3}\left(\frac{1}{2}\right)^3}=\frac{108n^2}{n(n-1)(n-2)}=\frac{108}{n}\frac{1}{\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)}\to0$$
