Teacher's proof of sequence convergence that I don't understand. Prove $a_n = \frac{n^2-2}{n^2 +2n + 2}$ converges to 1. 
Solution:
WTS $\exists L \in R, \forall \epsilon > 0, \exists N > 0$, such that for all $n \in N$, if $n > N$, then $|a_n - 1| < \epsilon$.
Let $\epsilon > 0$ be arbitrary
Choose N = $\max(\frac{6}{\epsilon}, 1) > 0$
Suppose n > N, then
$$\begin{align}
\left|\frac{n^2-2}{n^2+2n+2} - 1\right|&= \left|\frac{-2n-4}{n^2+2n+2}\right| 
\qquad\text{(By algebra)}\\
&=\frac{2|n+2|}{|n^2+2n+2|}       \qquad \text{ (abs value properties)}\\
&\leq \frac{2|n+2|}{n^2}          \qquad\qquad \text{ (By min denominator)}\\
&= 2\frac{1}{n^2} n \left|1+\frac{2}{n}\right|     \qquad \text{ (factoring)}\\
&= 2\frac{1}{n}  \left|1+\frac{2}{n}\right|        \qquad \text{ (factoring/potential problem)}
\end{align}$$
Helper Assumption:
If $n > 1$ ((ie) $ N \geq 1$)
$$\text{Note 1: } \max\left(\frac{6}{\epsilon}, 1\right) = \frac{1}{6} \implies \frac{1}{n} < 1$$
$$\begin{align}
\text{Note 2: } \max\left(\frac{6}{\epsilon}, 1\right)=1 &\implies\frac{2}{n}<2\\
&\implies 1 + \frac{2}{n} < 3
\end{align}$$
Thus:
$$|a_n-1|\leq \frac{6}{n} < \frac{6}{N} = \epsilon$$

This is the entire proof, I get everything until we get to the potential problem / helper assumption. Could someone explain, or give a easier way because I thought this question could have been much easier. 

Here's a much easier solution:
We have
$$1 > \frac{n^2 - 2}{n^2 + 2n + 2} > \frac{n^2 - 4}{n^2 + 4n + 4} = \frac{(n - 2)(n + 2)}{(n + 2)^2} = \frac{n - 2}{n + 2},$$
thus
$$\left\vert \frac{n^2 - 2}{n^2 + 2n + 2} - 1 \right\vert < \left\vert \frac{n - 2}{n + 2} - 1\right\vert,$$
so it suffices to show that
$$\lim_{n\to\infty}\frac{n-2}{n+2} = 1.$$
Pick $\epsilon > 0$ and choose $N > \tfrac{4}{\epsilon}$. Then if $n > N$, we have
$$\left\vert \frac{n - 2}{n + 2} - 1\right\vert = \left\vert \frac{-4}{n + 2}   \right\vert < \frac{4}{n} < \epsilon.$$
