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

• Seems quite right, though the final implication must be in the other direction.
– user65203
Aug 6, 2019 at 13:30
• It's wrong, for the reason Yves gave. Aug 6, 2019 at 13:34
• Did you finish this problem @user100000001? Sep 20, 2020 at 18:48

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

$$n\geq N > \frac{1}{64\epsilon} \implies n>\frac{1}{64\epsilon} \implies\epsilon > \frac{1}{64n}$$

Like the other commenters have mentioned, the implication is in the wrong direction. Since you want $$|\frac{1}{64n}| < \epsilon$$, you are required to have $$n > \frac{1}{64 \epsilon}$$. In other words, $$n > \frac{1}{64 \epsilon}$$ implies $$|\frac{1}{64n}| < \epsilon$$. What you have written is the other way around.

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.

Let $$\epsilon >0$$. Let $$N$$ be an integer greater than --insert your choice for $$N$$ here--.
If $$n \geq N$$, we have