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!

$$\begin{align*}
\frac5\epsilon &< \left|x+3\right|\\
\frac5\epsilon &< -\left(x+3\right)\\
x &< -\frac5\epsilon-3
\end{align*}$$
Hence, given $\epsilon>0$, we can take $M= -\frac5\epsilon-3$ such that $x<M$ implies $$\left|\frac{x+8}{x+3}-1\right|<\epsilon$$
