Proving limits using $\epsilon-\delta$ definition Would this be the proper way to prove a limit of a function using the $\epsilon-\delta$ definition? 

Use the definition of limit to show that $\lim_\limits{x\to2}x^2+4x=12$

Let $\epsilon>0$, then there exists $\delta>0$, such that $x\in \mathbb{R}$, $x\neq 2$ if
$$|x-2|<\delta\implies |x^2+4x-12|<\epsilon$$
$$|x^2+4x-12|=|x+6|\cdot|x-2|$$
Let $\delta = \frac {\epsilon}{|x+6|}$, then
$$|f(x)-12|<|x+6|\cdot \delta$$
$$|f(x)-12|<|x+6|\cdot \frac {\epsilon}{|x+6|}$$
$$|f(x)-12|<\epsilon$$
$$\therefore \lim f(x)=12$$
 A: No. Its not correct. $\delta$ cannot depend on $x$. This is one of those $\delta = min \{1, \epsilon /M\}$ problems. Use the fact that if $x \in (1,3)$ then $|x+6| <9 $. In particular, if $|x-2|< 1 $ then $|x+6| < 9$. Sometimes, when you can't factor or uniformly bound a term you have to settle for bounding it near the limit point. This is perfectly fine because all that matters when taking limits is the behavior of the function near the limit point anyways. If you set $\delta = min\{1, \epsilon /9\}$ , then $$|x-2| < \delta \implies |x-2||x+6|< \epsilon$$ 
EDIT:
Btw, the reason that delta cannot depend on x is because you have to find ONE delta that works for EVERY x. Having delta depend on x will give you a different delta for each value of x.
EDIT 2:
In response to your comment. You're welcome. I was crucified on my first analysis test because I was absent the day he went over this. So now I really harp on people when I see them do it. I can virtually guarantee you that your text has some worked examples of this type of problem (never have come across one that doesn't), and also that a question like this will be on the exam. Another way to see this is to pay attention to order of quantifiers.
$$(\forall \epsilon > 0)(\exists \delta > 0)(\forall x \in I)\big(0 < |x-a| < \delta \implies |f(x) - L| < \epsilon\big)$$
$ \\ $
Your approach would require the order to be $(\forall \epsilon > 0)(\forall x \in I)(\exists \delta > 0)$
Getting familiar with quantifiers will help you a lot in this class in both understanding subtle distinctions in definitions (like uniform continuity vs continuity) and in proofs by contradiction.
For example, how would one prove a limit doesn't exist? It's easy to see by using quantifier negation:
$$
\begin{align}
\sim (\forall \epsilon > 0)(\exists \delta > 0)(\forall x \in I)\big(0 < |x-a| < \delta \implies |f(x) - L| < \epsilon\big) \\
(\exists \epsilon> 0)\sim(\exists \delta > 0)(\forall x \in I)\big(0 < |x-a| < \delta \implies |f(x) - L| < \epsilon\big) \\
(\exists \epsilon > 0)(\forall \delta > 0)\sim(\forall x \in I)\big(0 < |x-a| < \delta \implies |f(x) - L| < \epsilon\big) \\
(\exists \epsilon > 0)(\forall \delta > 0)(\exists x \in I)\sim\big(0 < |x-a| < \delta \implies |f(x) - L| < \epsilon\big) \\
(\exists \epsilon > 0)(\forall \delta > 0)(\exists x \in I)\big(0 < |x-a| < \delta \, \land \, |f(x) - L| \geq \epsilon\big) \\
\end{align}
$$
Thus to show a limit doesn't exist you have to find an $\epsilon$ such that for every $\delta$, one can find an $x$ with the following property
$$|x-a| < \delta \ \text{ but } \ |f(x) - L| \geq \epsilon$$
A: After this step $|f(x)−12|<|x+6|⋅δ$, you want to try to set a bound on $|x+6|$.
We know $|x-2|<\delta$.
$\implies$ $-\delta<x-2<\delta$. 
$\implies$ $8-\delta<x+6<8+\delta$. 
We also know for obvious reasons, $-8-\delta<8-\delta$. 
$\implies -8-\delta<x+6<8+\delta$
$\implies |x+6|<\delta+8$. 
This lets us say, $|f(x)−12|<|x+6|⋅δ<(\delta+8)⋅\delta$
We can choose $\delta$ to be whatever we want, to avoid dealing with $\delta^2$, fix $\delta\leq1$.
Now $(\delta+8)⋅\delta\leq9\delta$
So pick $\delta=min\{1,\frac{\epsilon}{9}\}$
