<s>[INST] Finishing my $\epsilon - \delta$ proof Hi fellas I have this limit: $$lim_{x\to-1}\dfrac{x^2-1}{x^2+x}=2$$
$Dem:$
Let $\epsilon >0$ random but fixed then:
$$|\dfrac{x^2-1}{x^2+x}-2|=|\dfrac{x^2-1-2x^2-x}{x^2+x}|=|\dfrac{-x^2-2x-1}{x^2+x}|=|\dfrac{(x+1)^2}{x(x+1)}|=|\dfrac{x+1}{x}|<M|x-1|<M(\dfrac{\epsilon}{M})=\epsilon$$
So I have to choose M such that $|\dfrac{1}{x}|<M$  
Let's suppose that $|x+1|<c$ for some $c$. (My problem is that I don't know how to choose this $c$)
So $|x-1|<c \Rightarrow -c<x-1<c$ and so $-c-1<x<c-1$.  
The problem is that for a little $c<1$, $-c-1$ and $c-1 $ are both negative and for a bigger c, $-c-1$ is always negative and so I can't find my beloved $M$. 
 [/INST] What you want to show is that,
for any $\delta > 0$
there is an $\epsilon > 0$
such that
if
$|x+1| < \epsilon$
then
$|\dfrac{x^2-1}{x^2+x}
-2| < \delta
$.
You have worked out
very nicely that
$|\dfrac{x^2-1}{x^2+x}
-2| 
=|\frac{x+1}{x}|
$.
Suppose
$|x+1| < \epsilon$.
To make sure that
the $x$ in the denominator
does not cause problems,
we want to choose $x$ so it is
not close to zero.
If we choose
$\epsilon < \frac12$,
then,
since
$-1-\epsilon < x
< -1+\epsilon
$,
then
$x < -\frac12$,
so
$|x| > \frac12$.
Therefore,
$|\frac{x+1}{x}|
<\frac{\epsilon}{\frac12}
=2\epsilon
$.
This shows that
if
$|x+1| < \epsilon$
and
$\epsilon < \frac12$,
then
$|\dfrac{x^2-1}{x^2+x}
-2| < 2\epsilon
$.
Therefore,
to make
$|\dfrac{x^2-1}{x^2+x}
-2| < \delta
$,
choose
$2\epsilon
< \delta$,
or
$\epsilon
< \min(\frac12,\frac{\delta}{2})
$.
</s>