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