<s>[INST] 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.
 [/INST] $\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)}
$.
</s>