real analysis - Determine the limit and give and $epsilon-N$ proof of $lim_{nrightarrowinfty}frac{3+(-4)^n}{n^2}$.

I'm having difficulty working out this particular limit. My intuition is that it diverges because of the oscillating $(-4)^n$ which clearly grows much faster than then $n^2$ in the denominator. So in the limit the partial sequence would be oscillating between very large positive numbers and very large negative numbers. With that said my strategy would be to show that the limit of the reciprocal converges to $0$ since I have a theorem that says that a sequence diverges to infinity if and only if the reciprocal converges to $0$, so
\begin{align*}
\lim_{n\rightarrow\infty}\frac{n^2}{3+(-4)^n}=0.
\end{align*}
is what I'm working with. So far I have yet to come up with any promising avenue to construct an $\epsilon-N$ proof. Any ideas would be greatly appreciated.

Answer

How about showing $\lim_{m\to\infty}\frac{3+(-4)^{2m}}{(2m)^2}=\infty$?