Using the definition of the limit to show I am stuck an how to proceed in my proof. I am suppose to show, using the definition of the limit, that $\lim_{x \rightarrow 2}{(x^3-2x-4)/(x^2-4)=5/2}$.
I did some algebra with the $|f(x)-5/2|$ and got to the point where I have $|f(x)-5/2| \leq (1/2)|x-2||2x+3|$, but I am unsure how to procced in choosing my $\delta$... Any help would be appreciated. 
EDIT: Sorry, I should have been more clear; I need to do this using an epsilon and delta argument...
 A: First $$\begin{align}\left|f(x)-\frac{5}{2}\right|&=\left|\frac{x^2+2x+2}{x+2}-\frac{5}{2}\right|\\&=\left|\frac{x^2+2x+2}{x+2}-\frac{10}{x+2}+\frac{10}{x+2}-\frac{5}{2}\right|\\&\leq\left|\frac{x^2+2x+2}{x+2}-\frac{10}{x+2}\right|+\left|\frac{10}{x+2}-\frac{5}{2}\right|\\&=\frac{|x^2+2x+-8|}{|x+2|}+\frac{|10-5x|}{|2x+4|}\\&=\left(\frac{|x+4|}{|x+2|}+5\frac{1}{|2x+4|}\right)|x-2|\\&\leq\left(\frac{|x-2|+6}{4-|x-2|}+\frac{1}{8-2|x-2|}\right)|x-2|\\&\leq(\frac{6+\delta}{4-\delta}+\frac{1}{8-2\delta})\delta\\&\leq(7/3+1/6)\delta\end{align}$$
Now just put $\delta=\min(1/(7/3+1/6),\epsilon/(7/3+1/6))$.
Above we used things like triangle inequality several times. E.g. $$|x-2+4|\leq |x-2|+4.$$
Or as in $$|x+2|=|4-(-(x-2))|\geq|4|-|x-2|.$$ Or in $$|2x+4|=|2(x-2)+8|\geq 8-2|x-2|.$$

To proceed from where you got it.
You have $$\begin{align}|f(x)-5/2|&\leq(1/2)|x-2||2x+3|\\&=(1/2)|x-2||2(x-2)+7|\\&\leq(1/2)(2|x-2|+7)|x-2|\\&\leq(1/2)(2\delta+7)\delta\end{align}$$

A: Note: OP edited question after I've posted this solution.
$$\begin{align}\lim_{x \to2}\frac{x^3-2x-4}{x^2-4}&=\lim_{x \to2}\frac{(x-2)(x^2+2x+2)}{(x-2)(x+2)}\\
&=\lim_{x \to2}\frac{x^2+2x+2}{x+2}\\
&=\frac{2^2+2\cdot2+2}{2+2}\\
&=\frac{5}{2}
&\end{align}$$
