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 $\delta=\frac{\epsilon}{5}$

The whole point of epsilon-delta proofs is given any arbitrary range around $x=-3$, you want to show that you can get arbitrary close to your supposed limit as possible. 
Your range is within $\delta$ of $-3$ so:
$$0<|x-(-3)|<\delta$$
Now you just need to show that you can get arbitrary close to your supposed limit $-6$. In other words, you want to show that in your range the distance between between $f(x)$ and $L=-6$ is less than some arbitrary $\epsilon$:
So you want to show:
$$0<|x-(-3)|<\delta \implies 0<|(5x+9-(-6)|<\epsilon$$
Where 
$$A \implies B$$
Means "if $A$ then $B$"
Your proof is correct, but let me advice you that in order to understand it you need to know what arbitrary means. It means "not assigned a specific value".
In terms of your other question on how we go from $|(5x+9)-(-6)|=5|x+3|$ to $|(5x+9)-(-6)|<5\delta$ ,note this just follows from your given range that $0<|x+3|<\delta$ so multiplying both sides by $5$ gives $0<5|x-(-3)|<5\delta$. 
