calculus - Limit as n goes to infinity of a limit involving factorials and exponents

$$\lim_{n \to \infty} = \dfrac{3((n+1)!)(n-1)}{3^n + (n!)n^2}$$

I'm stuck with this limit. I managed to rewrite it to the following form:

$$\dfrac{3n-3}{\dfrac{3^n}{(n+1)!}+\dfrac{n^2}{(n+1)}}$$ but I don't know how to further simplify it. To me it seems like this goes to 0 because $3^n$ grows faster than any other term in the function, but wolfram alpha tells me it's $3$.

Answer

Divide the top and bottom by $n!$:

$$\lim_{n\to\infty} \dfrac{3((n+1)!)(n-1)}{3^n+(n!)n^2}=\lim_{n\to\infty} \dfrac{3(n+1)(n-1)}{\dfrac{3^n}{n!}+n^2}=\lim_{n\to\infty}\dfrac{3n^2-3}{n^2}=\lim_{n\to\infty}3-\dfrac{3}{n^2}=3$$
The thing is that $n!$ grows way faster than $3^n$ once $n>3$.