Formal Epsilon Argument Proof for Sequence 
Using a formal epsilon argument, show that $$\lim_{n \to \infty} \sqrt{n^2 + 2} - \sqrt{n^2 -2} =0 .$$

This is what I have, but i'm unsure if it's safe to assume that $\sqrt{n^2 - 2} >0$ or if I made any other errors.
Proof:
$$\left |\sqrt{n^2 + 2} - \sqrt{n^2 - 2} - 0\right | = \left(\sqrt{n^2 + 2} - \sqrt{n^2 - 2}\right)\cdot \frac{\sqrt{n^2 + 2}+\sqrt{n^2 - 2}}{\sqrt{n^2 + 2}+ \sqrt{n^2 - 2}}$$
$$\ \ = \frac{n^2 + 2 - n^2 + 2}{\sqrt{n^2 + 2}+\sqrt{n^2 - 2}}$$
$$\ \ = \frac{4}{\sqrt{n^2+2}+ \sqrt{n^2-2}}$$ (Because $\sqrt{n^2 - 2} > 0$)
$$< \frac{4}{\sqrt{n^2+2}} < \frac{2}{n} \ \ \ \ \ \ \ \ $$
We must find $N \in \mathbb{N}$ s.t. $\frac{2}{N} < \varepsilon \iff N > \frac{2}{\varepsilon}$
Fix $\varepsilon > 0$.
By the Archimedean Property, find $N \in \mathbb{N}$ with $N > \frac{2}{\varepsilon}$
Then if $n \geq N$,
$$|a_n-L| = \left |\sqrt{n^2 + 2} - \sqrt{n^2 - 2} - 0\right | < \frac{2}{n} \leq \frac{2}{N} < \varepsilon \ \ \  \text{(From Above)}$$
Thus proving, $$\lim_{n \to \infty} \sqrt{n^2 + 2} - \sqrt{n^2 -2} =0 .$$

It looks fine. 
To make sure $\sqrt{n^2-2}>0$, you can impose conditions such that $N \ge 2$ and $N \ge \frac2{\epsilon}$. 
For example, Let $N = 2 +  \lceil \frac2{\epsilon}\rceil .$
