Prove $\lim_{x\to-2}\frac{x+8}{x+3}=6 $ using $\epsilon-\delta$ 
Prove:
$$\lim_{x\to-2}\frac{x+8}{x+3}=6 $$

I started with:
$ |\frac{x+8}{x+3} - 6| < \epsilon  => |\frac{x+8-6x-18}{x+3} | < \epsilon => |\frac{-5x-10}{x+3} | < \epsilon =>| \frac{-5(x+2)}{x+3} | < \epsilon $
From the definition of limit we know that:
$ |x+2| <\delta $
Just don't know how to continue now... any suggestions?

Hint:
Take $\delta=min(\frac{1}{2},\frac{\epsilon}{10})>0$
Then $\forall x, |x+2| <\delta $, $|x+3|>\frac{1}{2}$, hence $$| \frac{-5(x+2)}{x+3} |\le 10|x+2| < \epsilon $$
The key is to select $\delta$ such that $x$ is closed to $-2$ compared with $-3$.
