Finding the limit of a sequence containing a factorial $\lim \limits_{n \to \infty}\frac{n^3-n^2-2}{n!}.$ I have the limit $$\lim \limits_{n \to \infty}\frac{n^3-n^2-2}{n!}.$$ I am able to convert it into the form $$\lim \limits_{n \to \infty}\frac{1}{\frac{n!}{n^3}}$$ but am stuck on how to solve $$\lim \limits_{n \to \infty}\frac{n!}{n^3}.$$ I know that $n!$ grows at a much larger rate than $n^3$ but am unsure how to express this in mathematical terms. 
 A: Hint
$$\frac{n^3-n^2-2}{n!}=\frac{1}{(n-3)!}+\frac{2}{(n-2)!}-\frac{2}{n!}$$
A: Another hint:
Note first that $\;u_n\sim_\infty\dfrac{n^3}{n!}$, so 
$$\frac{u_{n+1}}{u_n}\sim_\infty \biggl(\frac{n+1}n\biggr)^3\frac1{n+1}\sim_\infty \frac1{n+1}\to 0.$$
A: Note that for $n$ large the expression is positive and $n!>n^4$ (you can prove easily a part this two assumptions) then
$$0\le\frac{n^3-n^2-2}{n!}\le\frac{n^3-n^2-2}{n^4}=\frac1{n}-\frac1{n^2}-\frac2{n^4}\to 0$$
thus for squeeze theorem 
$$\lim \limits_{n \to \infty}\frac{n^3-n^2-2}{n!}=0$$
A: As other answers have noted, there are plenty of fancy ways to show the limit you want to show. But remember, you can always go back to the definition of a limit and work from there. If we wish to show that
$$\lim \limits_{n \to \infty}\frac{n!}{n^3}$$
goes to infinity, we must show that for any choice of lower bound $L$ there exists an integer $m$ such that $m<n$ implies that $L<\frac{n!}{n^3}$
We begin by noting that $1 < \frac{(n)(n-1)(n-2)(n-3)}{n^3}$ for all $n>6$.  (Can you prove this fact?)
Therefore, for our $m$ we can choose $L+4$ or $6$, whichever is larger, and we have
$$\frac{n!}{n^3} = (\frac{(n)(n-1)(n-2)(n-3)}{n^3})(n-4)(...) $$
if $n>m$ then we have the first part being greater than one, the $(n-4)$ part being greater than $L$, and the $(...)$ part being greater than one, so we can conclude that the whole product is greater than $L$, and we're done. 
Going back to the definition is tedious but it gets the job done, and it's good practice.
