$\epsilon - N$ 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 $\sqrt{4n^2+n} - 2n = \frac{n}{\sqrt{4n^2+n} + 2n}$. 
$$\left|\frac{n}{\sqrt{4n^2+n} + 2n} - \frac{1}{4}\right| \\
= \left|\frac{2n - \sqrt{4n^2+n}}{4(\sqrt{4n^2+n} + 2n)}\right|=\left|\frac{\sqrt{4n^2+n} - 2n}{4(\sqrt{4n^2+n} + 2n)}\right|\\
 = \left|\frac{n}{4(\sqrt{4n^2+n} + 2n)^2}\right| \leq \left|\frac{n}{4(4n)^2}\right| = \left|\frac{n}{64n^2}\right| \\
= \left|\frac{1}{64n}\right| < \epsilon \\
\implies n>\frac{1}{64\epsilon}$$

You did the scratch work correctly. But I wouldn't call this a proof (of course it contains all ingredients of a good proof!). 
You didn't introduce $\epsilon$. Of course, everyone knows what you mean but one should write it out if one wants to be fully rigorous. 
So, your proof should read:
Let $\epsilon >0$. Let $N$ be an integer greater than --insert your choice for $N$ here--.
If $n \geq N$, we have
--insert your estimations here--
Thus, by definition of limit, the result follows. QED
