Prove $ \lim_{x\to -\infty } \frac{x+8}{x+3} = 1 $ using only the definition of a limit

Question

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!

Answer 1

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

Answer 2

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 $ϵ$.