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.

Update If you really need to use a complete $\epsilon$-definition type proof, then using this method will only require that you additionally prove $\lim_{n\to\infty} \frac{n}{4n+1} = \frac{1}{4}$, which is considerably easier. (Just note that $\left|\frac{n}{4n+1}-\frac{1}{4}\right| = \frac{1}{4(4n+1)}$)
We will bound the terms eventually as follows:
For any $\epsilon > 0$ there exists some $N$ so that for all $n\geq N$: $$n^2+2 < n^2(1+\epsilon)$$
(i.e., if $2 < N^2\epsilon$), also clearly:
$$(1-\epsilon)n^2 < n^2+2$$
Putting these facts together, for $n\geq N$
$$\sqrt{1-\epsilon}\frac{n}{4n+1} = \frac{\sqrt{n^2(1-\epsilon)}}{4n+1} < \frac{\sqrt{n^2+2}}{4n+1} < \frac{\sqrt{n^2(1+\epsilon)}}{4n+1} = \sqrt{1+\epsilon}\frac{n}{4n+1} $$
Now taking limits we have:
$$
\lim_{n\to\infty} \sqrt{1-\epsilon}\frac{n}{4n+1} 
\leq \lim_{n\to\infty} \frac{n^2+2}{4n+1}
\leq \lim_{n\to\infty} \sqrt{1+\epsilon}\frac{n}{4n+1}
$$
so
$$\frac{\sqrt{1-\epsilon}}{4} \leq \lim_{n\to\infty} \frac{n^2+2}{4n+1} \leq \frac{\sqrt{1+\epsilon}}{4}$$
as this holds for all $\epsilon > 0$, let $\epsilon \to 0^+$:
$$\frac{1}{4} = \lim_{\epsilon\to 0^+}\frac{\sqrt{1-\epsilon}}{4} \leq \lim_{n\to\infty} \frac{n^2+2}{4n+1} \leq \lim_{\epsilon\to 0^+}\frac{\sqrt{1+\epsilon}}{4} = \frac{1}{4}$$
so the result follows from the sandwich theorem.
