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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

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    @MTurgeon thanks so much for all your help .. i posted an answer so maybe someone else can oneday can use it to help them understand too :)2012-06-01

1 Answers 1

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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 !