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 $M$.
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}) $.