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.
 A: 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.
A: Let $\epsilon>0$

$$\left|\frac {4\sqrt {n^2 +2} -4n-1}{16n+4}\right|\leq\frac{4n+8-4n-1}{16n+4}=\frac{7}{16n+4} \leq \frac{7}{16n}$$

We have that $\frac{7}{16n} \to 0$
Thus exists $n_0 \in \mathbb{N}$ such that $\frac{7}{16n}< \epsilon, \forall n \geq n_0$ 
So $\frac{1}{n}<\frac{16\epsilon}{7} \Rightarrow  n> \frac{7}{16\epsilon}$
Take $n_0=[\frac{7}{16\epsilon}]+1$ and we have that $$\forall n\geq n_0=[\frac{7}{16\epsilon}]+1 \Rightarrow \frac{7}{16n}<\epsilon \Rightarrow \left|\frac {4\sqrt {n^2 +2} -4n-1}{16n+4}\right| < \epsilon $$
Note that $[x]$ is the integer part of $x$
A: \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}
A: $\lim_{n \to \infty} \frac{\sqrt {n^2 +2}}{4n+1}=\frac14
$
$\begin{array}\\
\dfrac{\sqrt {n^2 +2}}{4n+1}
&\gt \dfrac{n}{4n+1}\\
&= \dfrac{n+1/4-1/4}{4n+1}\\
&=\dfrac14- \dfrac{1}{4(4n+1)}\\
\end{array}
$
so
$\dfrac{\sqrt {n^2 +2}}{4n+1}-\dfrac14
\gt - \dfrac{1}{4(4n+1)}
$.
Since
$(n+1/n)^2
=n^2+2+1/n^2
\gt n^2+2
$,
$\sqrt{n^2+2}
\lt n+1/n
$
so
$\begin{array}\\
\dfrac{\sqrt {n^2 +2}}{4n+1}-\dfrac14
&\lt \dfrac{n+1/n}{4n+1}-\dfrac14\\
&= \dfrac{4n+4/n-(4n+1)}{4(4n+1)}\\
&= \dfrac{4/n-1}{4(4n+1)}\\
&= \dfrac{1}{n(4n+1)}-\dfrac{1}{4(4n+1)}\\
&\lt 0
\qquad\text{for } n > 4\\
\end{array}
$
Therefore,
for $n > 4$,
$|\dfrac{\sqrt {n^2 +2}}{4n+1}-\dfrac14|
\lt \dfrac{1}{4(4n+1)}
$.
By choosing
$\dfrac{1}{4(4n+1)}
\lt \epsilon
$,
or
$n
\gt \dfrac14(\dfrac1{4\epsilon}-1)
$,
the difference is less then $\epsilon$.
Two notes:
Choosing
$n \gt \dfrac1{16\epsilon}$
is sufficient.
From the above,
$0 
\lt \dfrac{\sqrt {n^2 +2}}{4n+1}-\dfrac14+\dfrac{1}{4(4n+1)}
\lt \dfrac{1}{n(4n+1)}
$.
A: If you can use standard limit theorems about continuous functions, factor $n$ out of the numerator and the denominator (using fractions in both top and bottom), cancel them, etc.
A: 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.
