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?

Just factorise:
$$\frac{x^2+x}{x^2-x-2} \equiv \frac{x(x+1)}{(x-2)(x+1)}$$
Assuming $x \neq -1$, we can divide the numerator and denominator by $x+1$. For all $x \neq -1$, we have
$$\frac{x^2+x}{x^2-x-2} = \frac{x}{x-2}$$
As $x \to -1$, we see that 
$$\frac{x^2+x}{x^2-x-2} \to \frac{-1}{-1-2} = \frac{1}{3}$$
