How to prove that $\lim_{n \to \infty} \frac{\sqrt {n^2 +2}}{4n+1}=\frac14$? 
How to prove that $\lim_{n \to \infty} \frac{\sqrt {n^2 +2}}{4n+1}=\frac14$?

I started my proof with Suppose $\epsilon > 0$ and $m>?$ because I plan to do scratch work and fill in.
I started with our conergence definition, i.e. $\lvert a_n - L \rvert < \epsilon$
So $\lvert \frac{\sqrt {n^2 +2}}{4n+1} - \frac {1}{4} \rvert$ simplifies to $\frac {4\sqrt {n^2 +2} -4n-1}{16n+4}$
Now $\frac {4\sqrt {n^2 +2} -4n-1}{16n+4} < \epsilon$ is
simplified to $\frac {4\sqrt {n^2 +2}}{16n} < \epsilon$ Then I would square everything to remove the square root and simplify fractions but I end up with $n> \sqrt{\frac{1}{8(\epsilon^2-\frac{1}{16}}}$
We can't assume $\epsilon > \frac{1}{4}$ so somewhere I went wrong. Any help would be appreciated.

\begin{align}
\lim_{n \rightarrow \infty} \dfrac{\sqrt{n^{2} + 2}}{4n + 1} = \lim_{n \rightarrow \infty} \dfrac{\frac{1}{n}\sqrt{n^{2} + 2}}{\frac{1}{n}(4n + 1)} = \lim_{n \rightarrow \infty} \dfrac{\sqrt{1 + \frac{2}{n^{2}}}}{\left(4 + \frac{1}{n} \right)} = \dfrac{1}{4}
\end{align}
