Calculate $\lim_{n\to\infty}\frac{5n^n}{3n!+3^n}$ Could I please have a hint for finding the following limit?$$\lim_{n\to\infty}\frac{5n^n}{3n!+3^n}$$
 A: Stirlings approximation of the factorial:
$$
\sqrt{2\,\pi\,n}\left(\frac{n}{e}\right)^n
$$
5 and 3 are constant factors so put them before the limit. Now we have to concern ourselves with:
$$
\frac{5}{3}\lim_{n\rightarrow\infty}\dfrac{n^n}{\sqrt{2\,\pi\,n}\left(\frac{n}{e}\right)^n+3^n} = \frac{5}{3\,\sqrt{2\,\pi}}\lim_{n\rightarrow\infty}\dfrac{n^n}{\sqrt{n}\frac{n^n}{e^n}+3^n} = \frac{5}{3\,\sqrt{2\,\pi}}\lim_{n\rightarrow\infty}\dfrac{1}{\dfrac{\sqrt{n}}{e^n}+\left(\frac{3}{n}\right)^n}\\
= \frac{5}{3\,\sqrt{2\,\pi}}\lim_{n\rightarrow\infty}{\dfrac{e^n}{\sqrt{n}}}
$$
A: I think the limit diverges as consider the inverse of limit i.e 
$$\lim_{n\to\infty} \frac{3n! + 3^n}{5n^n}$$
$$\lim_{n\to\infty} \frac{3}{5}*\frac{3^n}{n^n}$$ tends to zero as 3/n tends to zero as n tends to infinity.
Also $$\lim_{n\to\infty} \frac{3}{5}*\frac{n!}{n^n}$$ also tends to zero as $$ e\left(\frac{n}{e}\right)^n \leqslant n! \leqslant ne\left(\frac{n}{e}\right)^n$$ and using sandwich theorem on this we get the total limit tends to zero so the actual limit given diverges.
A: Use equivalents:
First $3n!+3^n=3n!+o(n!)$ so: $$ 3n!+3^n\sim_\infty3n!\sim_\infty 3\sqrt{2\pi n}\Bigl(\frac n{\mathrm e}\Bigr)^n$$
and
$$\frac{5n^n}{3n!+3^n}\sim_\infty\frac{\mathrm e^n}{3\sqrt{2\pi n}}\xrightarrow[n\to\infty]{}\infty.$$
