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?

$$
\begin{align}
x^2-x-2
&=x^2-2x\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2-2\\
&=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}-\frac{2\cdot4}{4}\\
&=\left(x-\frac{1}{2}\right)^2-\frac{9}{4}\\
&=\left(x-\frac{1}{2}\right)^2-\left(\frac{3}{2}\right)^2\\
&=\left(x-\frac{1}{2}-\frac{3}{2}\right)\left(x-\frac{1}{2}+\frac{3}{2}\right)\\
&=(x-2)(x+1)
\end{align}
$$
Now, things are going to cancel out nicely:
$$
\lim\limits_{x \rightarrow -1}\frac {x^2+x} {x^2-x-2}=
\lim\limits_{x \rightarrow -1}\frac {x(x+1)} {(x-2)(x+1)}=
\lim\limits_{x \rightarrow -1}\frac {x} {x-2}=\frac{-1}{-1-2}=\frac{-1}{-3}=\frac{1}{3}.
$$
