> 
> Let $f(x)=5x+9$. Show that $\lim \limits\_{x \to -3}f(x)=-6$
> 
> 
> 

A couple of questions about showing this and proving this. As I'm working through the problem I don't understand how I proved or showed anything as I don't understand the results I get.

I understand that $\delta$ is some range between $x$ and $L$, and $\epsilon$ is some range between $f(x)$ and $L$, so what do these proofs tell me?

An easier question, though, is while proving this, a term appears (my instructors notes) and I don't know where it came from:

We want to find $\delta$ such that $\left|f(x)-(-6)\right|=\left|(5x+9)-(-6)\right|\lt \epsilon$ whenever $0 \lt \left|x-(-3)\right| \lt \delta$

But $\left|(5x+9)-(-6)\right|=\left|5x+15\right|=5\left|x+3\right|=5\left|x-(-3)\right|\lt 5 \delta$ **(where does the $5\delta$ come from?)**. So, we want $5\delta=\epsilon$, which implies that $\delta=\frac{\epsilon}{5}$