Find the limit of $\frac {x^2+x} {x^2-x-2}$ where $x \to -1$? I need to find the limit of $\frac {x^2+x} {x^2-x-2}$ where $x \to -1$. Right now I am getting $\frac{0}{0}$ if I don't factor first, or $\frac{2}{0}$ if I do.
Here are my factoring steps:
$\frac {x^2+x} {x^2-x-2}$
$=\frac{x(x+1)}{(x-2)(x+1)}$
replace $x$ with $-1$
$=\frac{-1(-1+1)}{(-1-2)(-1+1)}$
$=\frac{2}{-3 (0)}$
$=\frac{2}{0}$
How can I solve this problem?

$$\frac{x^2+x}{x^2-x-2}=\frac{x(x+1)}{(x-2)(x+1)}=\frac x{x-2}\xrightarrow[x\to-1]{}\frac{-1}{-3}=\frac13$$
The above is justified by the fact that taking the limit when $\;x\to-1\;$ means $\;x\;$ gets closer and closer to $\;-1\;$ but never equals it in this limit process.
To calculate a limit as $\;x\to a\;$  is the same as substituting $\;x=a\;$ in the function iff the function is continuous at $\;a\;$ , otherwise it may fail...as in this case, where the function's not even defined at $\;x=-1\;$ 
