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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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    What you have to show is that **for a fixed $\epsilon>0$**, there exists a $x_0$ such that whenever $x\leq x_0$ the relation $\left|\frac{x^2+1}{x^2-1} -1\right|<\epsilon$ holds, and this has to be true **for all $\epsilon>0$**.2012-06-01
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    And so, if you want to be rigorous, you have to fix $\epsilon>0$, and find a $x_0$ (which might depend on $\epsilon$)2012-06-01
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    Ronald, welcome to math.SE, and let me express my appreciation that you are explaining what you've thought about your question so far - it is far more satisfying for people to help someone who has tried to do it themselves first! Keep it up!2012-06-01
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    @MTurgeon .. Thanks for that, I get what you mean but for some reason the only x0 i can find fits x>=x0.. which makes no sense but probably a silly mistake from rushing @ZevChonoles♦ yeah ive been loving this site since I found it .. only now have i ever actually make a post.. I get what you mean aswell, it must be seriously frustrating trying to help someone when you dont know what they do/dont understand2012-06-01
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    @Ronald Well, you are pretty close. At some point, you have figured out that $\left|\frac{x^2+1}{x^2-1}-1\right|=\frac{2}{|x^2-1|}$, and so you want $x$ such that $|x^2-1|>\frac{2}{\epsilon}$. I leave you the rest.2012-06-01
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    @MTurgeon aaaaaah you are a champion !2012-06-01
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    @Ronald I am glad I could be useful! If you wish, you can post your solution one it is complete and accept it. This way, your question won't go unanswered.2012-06-01
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    @MTurgeon ok wait you are still a champion, but one quick thing (just clarification).. so at the step x^2 > 2/ϵ-1 , the logical next step is x > sqrt(2/ϵ-1), but x is approaching negative infinity and so we want it to be in the form x<=f(ϵ)... so can we say that due to the nature of the square(always positive) that x > sqrt(2/ϵ-1) OR x < -sqrt(2/ϵ-1) ??2012-06-01
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    @Ronald When you take the square root, you have to keep the absolute value: $$x^2>2/\epsilon-1 \Rightarrow |x|>\sqrt{2/\epsilon -1}.$$2012-06-01
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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

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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 !