Let $f(x) = 5x+9$. Show that $\lim \limits_{x \to -3}f(x)=-6$ 
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 $\delta=\frac{\epsilon}{5}$

You want $\left|f(x)-(-6)\right|=5\left|x-(-3)\right|< \epsilon$, that is $\left|x-(-3)\right|\lt \epsilon/5$ and that is guaranteed if you take $\delta=\epsilon/5$.
