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.

For one, we have
$$
\frac{\sqrt{n^2+2}}{4n+1} \le \frac{\sqrt{n^2+2}}{4n} = \sqrt{\frac{1}{16}+\frac{1}{8n}} \le \frac{1}{4}+ \sqrt{\frac{1}{8n}},
$$
and we also have
$$
\frac{\sqrt{n^2+2}}{4n+1} \ge \frac{n}{4n+1} = \frac{1}{4 + 1/n}.
$$
As $n\to\infty$, both of these tend to $1/4$. By the squeeze theorem, we obtain the desired result.

If you need a more formal proof, you can go the extra mile like this:
Let $\epsilon>0$ be arbitrary. By the Archimedean property of $\Bbb R$, there exists $N_1\in\Bbb N$ such that if $n\ge N_1$, then
$\sqrt{\dfrac{1}{8n}}<\epsilon$. Thus $\lim_{n\to\infty}\dfrac{1}{4}+\sqrt{\dfrac{1}{8n}} = \dfrac{1}{4}$.
Now consider the difference 
$$
\left|\frac{1}{4}-\frac{1}{4+1/n}\right| = \frac{1}{4}\cdot\frac{1}{4n+1}.
$$
Again, by the Archimedean property of $\Bbb R$, there exists $N_2\in\Bbb N$ such that if $n\ge N_2$, then $\frac{1}{4}\cdot\frac{1}{4n+1} <\epsilon$. Thus $\lim_{n\to\infty}\dfrac{1}{4+1/n} = \dfrac{1}{4}$. By the squeeze theorem, the claim is proved.
