Prove $ \lim_{x\to -\infty } \frac{x+8}{x+3} = 1 $ using only the definition of a limit I need to prove this limit:
$$ \lim_{x\to -\infty } \frac{x+8}{x+3} = 1 $$
I started with:
$|\frac{x+8}{x+3}-1|< ϵ $

$|\frac{x+8-x-3}{x+3}|=|\frac{5}{x+3}|=\frac{5}{|x+3|} <ϵ$

$\frac{5}{\epsilon} < |x+3|$

How do I proceed from here? how the hell can I extract the expression inside the absolute value? Am I in the wrong direction? 
Thanks a lot!

Hint:
(And sorry I'm on my phone)
Given any $ϵ>0$, we want to find $M$ such that
$$
\left|\frac {x+8}{x+3}−1\right|<ϵ
$$
whenever $x<M$.
You have basically already done this.
Try to pick a $M$ that is a function of $ϵ$.
