proof limit of $\frac{1}{(x^{2}-1)}$ = -1 using $\epsilon - \delta$ as $x_{0} \to 0$ Question: proof limit of $\frac{1}{(x^{2}-1)}$ = -1 using $\epsilon - \delta$ as $x_{0} \to 0$
Hi... I am stumped on an intro analysis problem.
so it is stated Show $lim \frac{1}{x^{2}-1}$ = -1 as $x_{0} \to 0$. 
here is my work, and I can't quite simplify the $|f(x) - L | < \epsilon$ to get an expression for $\delta$.
|$\frac{1}{x^{2}-1} + 1| \leq |\frac{1}{(x+1)(x-1)} + 1 | \leq  | \frac{1}{x-1} - \frac{1}{x+1}  + 2| \leq \epsilon $    not sure if I should use partial fractions here.
I am not sure the neighborhood to select for $x_0$  , so I'm guessing $|x| < 1$
this choice yields  $1 < \frac{1}{x+1} < \frac{1}{x} < \frac{1}{x-1} $
so the expression of $\epsilon$ can be written: 
$|\frac{1}{x-1} - \frac{1}{x+1} + 2 | < |\frac{2}{x-1} + 2 | $ this can be simplified to 
$| x - 0 | < \frac{\epsilon}{2}(x-1)$   where given our choice of $x_{0}$ (x-1) $< 2$
so I chose $\delta = min( 1, \epsilon)$.
but I am not sure this work is any good, and am having trouble writing the proof.
sincerely thank you!
Here is the Proof  
let $\delta = min\{\frac{1}{2},\frac{5\epsilon}{2} \}$
let $\delta = \frac{1}{2} \implies x^{2} < \frac{|x|}{2} <\frac{1}{4}$
where $|x| < \frac{1}{2}$
then $\frac{-5}{4} < x^{2} - 1 < \frac{-3}{4} \implies \frac{5}{4} < |x^{2}-1| < \frac{3}{4} \implies \frac{1}{x^{2}-1}<\frac{4}{5}$
then $\frac{x^{2}}{x^{2}-1} < \frac{4|x|}{2*5} < \frac{4*1}{4*5} = \frac{1}{5} < \epsilon$
where if $\delta = \frac{1}{2} \implies \frac{1}{2} < \frac{5\epsilon}{2} \implies \frac{1}{5}<\epsilon$
for $\delta = \frac{5\epsilon}{2} $
let $|x - 0| < \frac{5\epsilon}{2} $ and for $|\frac{x}{2}|<\frac{5 \epsilon}{4}$ then
$x^{2}<|\frac{x}{2}| \implies |\frac{x^{2}}{x^{2}-1}| \leq \frac{2|x|}{5}$ 
so  $x^{2} < \frac{|x|}{2} < \frac{5\epsilon}{4}$
then
 $|\frac{x^{2}}{x^{2}-1}| < \frac{4|x|}{5}  <\frac{4}{5}*\frac{5\epsilon}{4} = \epsilon$ 
 A: Note that in our case,
$$|f(x)-L|=\frac{x^2}{|x^2-1|}.$$
If $|x|\lt 1/2$, then $|x^2-1|\gt \frac{3}{4}$ and therefore
$$|f(x)-L|\le \frac{4x^2}{3}.$$
Now it should not be hard to find an appropriate $\delta$. To make things even smoother, note that if $|x|\lt \frac{1}{2}$ then $x^2\le \frac{|x|}{2}$. 
Added: We wanted to show that given $\epsilon\gt 0$, we can find a $\delta$ such that if $|x-0|\lt \delta$, then $\frac{x^2}{|x^2-1|}\lt \epsilon$. The "top" can be easily made  small by choosing $|x|$ small enough. But the bottom could spoil things, because $\frac{1}{|x^2-1|}$ can be very large if $x$ is close to $1$ or $-1$. So the first thing to do is to neutralize the bottom by specifying that $\delta$ will be $\le 1/2$. Instead of $1/2$ we could have chosen $1/10$, or $8/10$.
We concluded that if $|x|\le 1/2$ then the expression we are trying to make small is $\le 4x^2/3$, which, since $|x|\le 1/2$, is $\le 2|x|/3$. This in turn is $\le |x|$. So if we make $\delta=\min(1/2,\epsilon)$, then if $|x|\lt \delta$ we will have $\frac{x^2}{|x^2-1|}\lt \epsilon$.
Note that we expended no effort whatsoever to choose the "largest" $\delta$ that would do the job. Choosing an optimal or near optimal $\delta$ might be of great importance if we were doing numerical analysis, but is of no importance here.   
A: Note that $$|f(x)-L|=\left|\frac1{x^2-1}-1 \right|=\left|\frac{x^2}{(x-1)(x+1)} \right|=\frac{x^2}{|x-1||x+1|}, $$ so if $|x|<\frac12$, then $|x-1||x+1|>2|x|$ and $|f(x)-L|<\frac12|x|$. So choosing $\delta=\min\left\{\varepsilon,\frac12\right\}$ we see that $|x|<\delta$ implies $$|f(x)-L|<\frac12|x|<\varepsilon. $$
