Elementary Real Analysis - Let me know if I'm on the right track So I'm guessing this is a pretty simple example for this topic but I just want to check myself as I'm new to this analysis area and not sure that what I'm saying is mathematically sound..
The question is show that $\lim\limits_{x \to -\infty}\frac{x^2+1}{x^2-1} =1$.
So we must show as $x \to -\infty$, $\left|\tfrac{x^2+1}{x^2-1}  -1\right| < \epsilon$.
My attempt is : 
\begin{align*}
\left|\dfrac{x^2+1}{x^2-1}  -1\right| &=\left|\dfrac{x^2+1}{x^2-1}  - \dfrac{x^2-1}{x^2-1} \right| \\\\
&=\left|\dfrac{(x^2+1)-(x^2-1)}{x^2-1} \right| \\\\
&=\left|\dfrac{x^2-x^2+1+1}{x^2-1}\right| \\\\
&=\left|\dfrac{2}{x^2-1}\right| < \left|\dfrac{2}{x^2-4}\right| = \left|\dfrac{2}{(x+2)(x-2)}\right| \\
\end{align*}
Now is where I'm not 100 percent sure that what I'm doing is right, can we then say that as $x$ approaches negative infinity, $(x+2)$ and $(x-2)$ become very large and negative, and therefore $2/(x+2)(x-2)$ becomes smaller and smaller and so, for any $\epsilon>0$, 
$$\epsilon > \left|\dfrac{2}{(x+2)(x-2)}\right| > \left|\dfrac{2}{x^2-1}\right| = \left|\dfrac{x^2+1}{x^2-1}  -1\right|$$   thus proving the original problem ... is this ok/rigorous? 
please dont be too hard on me I'm really still just trying to grasp the ideas and understand exactly what we are 'allowed' to do .. thanks for any help

Assume $\epsilon>0$, then 
$$\begin{align*} \left|\frac{x^2+1}{x^2-1} - 1\right| < \epsilon
&\Rightarrow \left|\frac{2}{x^2-1}\right| < \epsilon \\
&\Rightarrow \left|\frac{x^2-1}{2}\right| > 1/\epsilon \\
&\Rightarrow |x^2-1| > \frac{2}{\epsilon}. \end{align*}
$$
when $x^2-1 > 0$, $x^2>1 $, $|x| > 1$,
so, $(x^2-1)>2/\epsilon$ .... (when $|x|>1$)
$x^2 > 2/\epsilon+1$
$|x| > \sqrt{2/\epsilon+1}$ 
therefore :
Given any $\epsilon > 0 $, $\left|\frac{x^2+1}{x^2-1} - 1\right| < \epsilon$
when $|x| > \sqrt{2/ϵ+1}$.
ta da !
references : M Turgeon that king !
