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 $3$.
 A: 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 $n>3$.
A: First, compute $$\lim_{x \to \infty}\frac{3^n}{(n+1)!} =0.$$
With that in mind, divide the numerator and denominator of your expression by $n$ to get 
$$\lim_{x \to \infty} \frac{ 3-\frac{3}{n} }{ \frac{1}{n}\frac{3^n}{(1+n)!}+\frac{n}{n+1}}.$$
Then the numerator goes to $3$ and the denominator goes to $0+1$. 
A: From the step you are at, all you need to do is to show $\lim\limits_{n\to\infty}\frac{3^n}{n!}=0$.  It becomes clear via the following:
$$\frac{3^n}{n!}=\frac{\overbrace{3\times3\times3\times3\times\dots\times3}^n}{\underbrace{1\times2\times3\times4\times\dots\times n}_n}$$
Clearly after $n=3$, the factorial starts to overtake $3^n$, growing much much faster.
We may use the following inequality to show the limit is $0$:
$$\frac{3^n}{n!}=\frac{\overbrace{3\times3\times3\times3\times\dots\times3}^n}{\underbrace{1\times2\times3\times4\times\dots\times n}_n}<\frac3n\tag{very large $n$}$$
And see that
$$\lim_{n\to\infty}\frac{3^n}{n!}=0$$
A: Use equivalents and asymptotic analysis:
In the denominator, $3^n=_\infty o(n!)$ hence $3^n+(n!)n^2\sim_\infty (n!)n^2$.
In the numerator, $3\bigl((n+1)!\bigr)(n-1)\sim_\infty3n\bigl((n+1)!\bigr)$.
Thus $$\frac{3\bigl((n+1)!\bigr)(n-1)}{3^n + (n!)n^2}\sim_\infty \frac{3n\bigl((n+1)!\bigr)}{(n!)n^2}=\frac{3(n+1)}{n}\sim_\infty 3.$$
