How to I solve this limit to -3 I need help solving this limit $\lim \limits_{x \to -3} \frac{4x+12}{3x^3-27x}$.
I know that I am suppose to factor the function and then plug in -3 to calculate the result.
$\lim \limits_{x \to -3} \frac{4(x+3)}{3x^3-27x}$. But don't know how to factor $3x^3-27x$. 
How do I factor $3x^3-27x$? I'm I on the right path?
Thanks!

$$
\begin{split}
\mathop {\lim }\limits_{x \to  - 3} \frac{{4x + 12}}
{{3{x^3} - 27x}} &= \mathop {\lim }\limits_{x \to  - 3} \frac{{4\left( {x + 3} \right)}}
{{3x\left( {{x^2} - 9} \right)}}\\ 
&= \mathop {\lim }\limits_{x \to  - 3} \frac{{4\left( {x + 3} \right)}}
{{3x\left( {x - 3} \right)\left( {x + 3} \right)}} \\
&= \mathop {\lim }\limits_{x \to  - 3} \frac{4}
{{3x\left( {x - 3} \right)}} \\
&= \frac{4}
{{3\cdot\left( { - 3} \right)\cdot\left( { - 3 - 3} \right)}} = \frac{2}
{{27}}
\end{split}
$$
