Find the limit $x\to1$ of : $\vert x^2+x-2\vert /(x^2-1)$ Substitution of $1$ in the equation gives us $ \frac 00$ (indeterminate form), and so we must find the limit some other way.
By breaking $\vert x^2+x-2\vert $ up into $\vert (x-1)(x+2)\vert $ and noting that $\vert x-1\vert  = -(x-1)$ when $x < 1 $  and $\vert x-1\vert  = (x-1)$ when $x > 1$ and that $\vert x+2\vert  = x + 2$ when $x > -2$ we deduce the following:
$$\lim\limits_{x\to1^-}= \frac{-(x-1)(x+2)}{(x-1)(x+1)}= -(3/2)$$ 
(as the $(x-1)$ terms cancel.
Using a similar argument for the limit as $x\to1^+$ (from the right) we  see that it equals $3/2$ (since $\vert x-1\vert  = (x - 1)$.) 
We conclude that the limit does not exist as the limits taken from the left side and right side (approaching $1$) do not equate (i.e. $-3/2$ does not equal $3/2$).
Limit calculators say the answer is $0$, however, I can't seem to figure out how this would be so.
 A: Though your proof is completely accurate, I would like to offer a slightly different method (with the bonus effect that this question will probably be removed from te unanswered que).  
Let's say $f(x)=\dfrac{\vert x^2+x-2\vert}{x^2-1}$. You have already argued quite well that $f(x)<0$ in $(1-\varepsilon,1)$ and $f(x)>0$ in $(1,1+\varepsilon)$ (at least for sufficiently small $\varepsilon>0$).  
Now suppose that  $\lim\limits_{x\to1}f(x)=L$ for some $L$. This would mean that $$\forall\,\varepsilon>0 \, \exists \delta >0: 0<\vert x-1\vert<\delta \implies \vert f(x)-L\vert <\varepsilon.$$ In words this would mean that  $L$ lies arbitrarily close to $f(x)$ whenever $x$ lies close enough to $1$. So $L$ lies arbitrarily close to both negative and positive numbers, meaning $L=0$ is the only option. (This is arguably not entirely trivial, however very easily proved through contradiction.)  
Now for every $\varepsilon>0$ we should have:
$$0<\vert x-1\vert<\delta\implies \left\vert\frac { x^2+x-2}{x^2-1}\right\vert=\left\vert\frac{x+2}{x+1}\right\vert<\varepsilon$$ for some $\delta>0$.  But now for $x=1+\frac\delta2$ we have that $$\dfrac{x+2}{x+1}=\dfrac{3+\frac\delta2}{2+\frac\delta2}>1.$$ This is a contradiction and thus the limit is not $L=0$. However $L=0$ was the only possibility we had for $\lim\limits_{x\to1}f(x)=L$, and thus the limit does not exist.
