Find limit with ε − δ arguments Determine $\lim_{x \to p} f(x)$  justifying your answer with $\varepsilon$−$\delta$ arguments.
$$f(x) = \frac{1}{x^2−1}$$
$p = 0$; 
I only managed so far:
$$\frac{x^2} {x^2-1} <  \varepsilon$$
How do i simplify this expression to make it       $x <\varepsilon$ ?
 A: You can intuitively see that $\lim\limits_{x\rightarrow 0}f(x) = -1$. In consequence, what you are looking for is to prove that for any $\varepsilon >0$ there exists a $\delta > 0$ such that
$$x\in (-\delta,\delta) \Rightarrow \left| -1 - f(x) \right| < \varepsilon.$$
The last inequality can be rewritten as $\left| -1 - \dfrac{1}{x^2-1} \right| < \varepsilon$, which in turn is $\left| - \dfrac{x^2}{x^2-1} \right| < \varepsilon$.
Note: when dealing with limits we consider $\varepsilon, \delta$ to be very small, close to 0. In consequence, we expect $x \approx 0 < 1$, hence $x^2-1 < 0$. Thus, our absolute value relation can be transformed as follows
$$\dfrac{x^2}{1-x^2} < \varepsilon \Leftrightarrow x^2 < \varepsilon-x^2\varepsilon \Leftrightarrow (1+\varepsilon)x^2 < \varepsilon.$$
Now, as $\varepsilon \approx 0 < 1$, we have $(1+\varepsilon)x^2 < 2x^2$. Thus, it is sufficient to find an $x$ such that $2x^2 < \varepsilon$.
So, if we consider $\delta = \sqrt{\varepsilon/2}$, we have for any $x\in (-\delta,\delta)$ that $x^2 < \dfrac{\varepsilon}{2}$, therefore $(1+\varepsilon)x^2 < \varepsilon$ and the limit can be proven.
Hope it helps and clarifies a bit the thought process! 
A: One thing I wish students would realize is that figuring out how to do a $\delta-\epsilon$ proof is working backwards and what you end up is rough draft of a proof that you use to make a final proof.  
In this case the final proof will go like this.
$\lim_{x\to 0} \frac 1{x^2 - 1} = -1$.
Proof:  For any $\epsilon > 0$ let $\delta = \min(\sqrt{\frac 34 \epsilon}, \frac 12)$.
If $|x-0| < \delta;x\ne 0$ then $|x| < \frac 12$ and $x^2 < \frac 14$ and $1> 1-x^2 > \frac 34$
But as $0<|x| < \delta$ we have 
$|x^2| < \delta^2 < \frac 34\epsilon < |1-x^2|\epsilon$.
So $\frac {|x^2|}{|1-x^2|}< \epsilon$ 
And $|\frac 1{x^2-1}-(-1)|=|\frac {x^2}{x^2-1}|=\frac {|x^2|}{|1-x^2|}< \epsilon$.
So $\lim_{x\to 0} \frac 1{x^2 - 1} = -1$.
....
So the question is how the heck did we figure out to let $\delta = \min(\sqrt{\frac 43 \epsilon}, \frac 12)$ and how did we figure to do the manipulations.
ANd that was:
Firs $\frac 1{0^2 -1} =-1$ so we figure the limit has to be $-1$.
We want to end with $|\frac 1{x^2 -1} -(-1)| < \epsilon$.
And that is $|\frac 1{x^2 -1} -(-1)|=|\frac {x^2}{x^2-1}| < \epsilon$ so
$|x^2| < |x^2 - 1|\epsilon$.
Now we can make $|x-0| < \delta$ as small as we like so if make it really small then $|x^2 -1|$ can be really close to $1$.  If we, for example make $\delta \le \frac 12$ then $|x^2 -1| \ge \frac 34$ so we can get
$|x^2| < |x^2 - 1| \epsilon \le \frac 34 \epsilon$.
So If we have $|x^2| < \delta^2 \le \frac 34 \epsilon$
then it will work.  
So if we have $\delta \le \frac 12$ and $\delta \le \sqrt{\frac 34\epsilon}$ it will work.  Thus our chose of $\delta = \min(\frac 12,\sqrt{\frac 34\epsilon}$.  And our final proof.
