<s>[INST] $\epsilon - N$ proof of $\sqrt{4n^2+n} - 2n \rightarrow \frac{1}{4}$ I have the following proof for $\lim_{n\rightarrow\infty} \sqrt{4n^2+n} - 2n = \frac{1}{4}$ and was wondering if it was correct. Note that $\sqrt{4n^2+n} - 2n = \frac{n}{\sqrt{4n^2+n} + 2n}$. 
$$\left|\frac{n}{\sqrt{4n^2+n} + 2n} - \frac{1}{4}\right| \\
= \left|\frac{2n - \sqrt{4n^2+n}}{4(\sqrt{4n^2+n} + 2n)}\right|=\left|\frac{\sqrt{4n^2+n} - 2n}{4(\sqrt{4n^2+n} + 2n)}\right|\\
 = \left|\frac{n}{4(\sqrt{4n^2+n} + 2n)^2}\right| \leq \left|\frac{n}{4(4n)^2}\right| = \left|\frac{n}{64n^2}\right| \\
= \left|\frac{1}{64n}\right| < \epsilon \\
\implies n>\frac{1}{64\epsilon}$$
 [/INST] You want to show that $\lim_{n\rightarrow\infty} \sqrt{4n^2+n} - 2n = \frac{1}{4}$. To do this, I would split up the analysis into scratch work and the formal proof.
For the scratch work, you need to find a suitable upper bound. You have done this by showing
\begin{align}
\left|\frac{n}{\sqrt{4n^2+n} + 2n} - \frac{1}{4}\right| &
= \left|\frac{2n - \sqrt{4n^2+n}}{4(\sqrt{4n^2+n} + 2n)}\right|\\&=\left|\frac{\sqrt{4n^2+n} - 2n}{4(\sqrt{4n^2+n} + 2n)}\right|\\&
 = \left|\frac{n}{4(\sqrt{4n^2+n} + 2n)^2}\right| \\&\leq \left|\frac{n}{4(4n)^2}\right| \\&= \left|\frac{n}{64n^2}\right| \\&
= \left|\frac{1}{64n}\right| \\&< \epsilon 
\end{align}
which means that $n>\frac{1}{64\epsilon}$ is the upper bound.
For the formal proof:
Let $\epsilon>0$ (you need to fix $\epsilon$ as a small positive constant). It follows from $\frac{1}{\epsilon}>0$ that $\frac{1}{64\epsilon}>0$. Then by the Archimedean property there exists a $N\in\mathbb N$ such that $N>\frac{1}{64\epsilon}$. Then if $n\geq N > \frac{1}{64\epsilon}$, we have
$$\left|\frac{n}{\sqrt{4n^2+n} + 2n} - \frac{1}{4}\right|\leq \frac{1}{64n}<\epsilon $$
where the last inequality follows from
$$n\geq N > \frac{1}{64\epsilon} \implies n>\frac{1}{64\epsilon} \implies\epsilon > \frac{1}{64n}$$
</s>