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?

Note that:
$$n!\ge1\cdot 2\cdot3\cdot3\cdots\cdot3\cdot(n-2)\cdot(n-2)\cdot(n-2)=2\cdot3^{n-5}(n-2)^3$$
So for $n>5$:
$$\frac{2^nn^2}{n!}\le\frac{2^nn^2}{2\cdot3^{n-5}(n-2)^3}=\frac{3^5}{2}\left(\frac{2}{3}\right)^n\left(\frac{n}{n-2}\right)^3\frac{1}{n}<\frac{3^5}{2}\left(\frac{2}{3}\right)^5\left(\frac{5}{3}\right)^3\frac{1}{n}\to0$$
