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?

First let us restrict $x$ such that $\vert x+2 \vert < 1/2$, i.e., $-5/2<x<-3/2$. 
This means we have $1/2<x+3<3/2$, i.e., $\dfrac23 < \dfrac1{x+3} < 2$
Hence, we have
$$\left \vert \dfrac{-5(x+2)}{x+3}\right \vert = 5 \left \vert \dfrac{x+2}{x+3}\right \vert < 10 \vert x+2 \vert$$
Hence, choose your $\delta = \min(1/2,\epsilon/10)$ to ensure that $\left \vert \dfrac{-5(x+2)}{x+3}\right \vert$ is less than $\epsilon$.
