<s>[INST] 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
 [/INST] 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 !
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