Come back again with Limits!

Let $\displaystyle x=2-\delta$ where $\displaystyle \delta$ is a positive real number. then the limit is $\displaystyle \frac{|x^{2}-x-2|}{2-(x^{2}+4)^{1/3}}$ $\displaystyle \frac{|x-2||x+1|}{2-(8-4\delta)^{1/3}}=\frac{|x-2||x+1|}{2-2(1-\frac{1}{6}\delta)}=\frac{\delta|x+1|}{\frac{\delta}{3}}$

$\displaystyle \lim_{x->2^{-}}=3|x+1|=9$