how to demonstrate that $\lim_{x\to2}3x^2=12$ using $\gamma$ and $\epsilon$ definition how to demonstrate that $\lim_{x\to2}3x^2=12$ using $\gamma$ and $\epsilon$ definition? 
My current steps I have in order from top to bottom:
$|3x^2-12|<\epsilon$ and $0<|x-2|<\gamma$
assume $z=x-2$
$0<|z|<\gamma$ and $3z^2+12|z|<\epsilon$
$3z^2+12|z|<3\gamma^2+12\gamma=3\gamma(4+\gamma)$
Then assume $\gamma<1$
$\gamma=15$ hence $\gamma=\frac{\epsilon}{15}$
then now i am stuck and have no clue how to proceed
 A: Okay,  let's just say that everything you've done is scratch paper.  You have reached the conclusion $\delta$ should be $ \frac {\epsilon}{15}$ and $\delta < 1$.  So know we need to turn this into a proof.
For any $\epsilon > 0$, we can define $\delta = \min (1, \frac {\epsilon}{15})$.
Now for any $x$  so that $|x-2| < \delta$ (that is to say for any $x \in (2-\delta, 2+ \delta)$ then
$-\delta < x- 2< \delta$
$0 < 2 - \delta < x < 2 + \delta$
$(2-\delta)^2 < x^2 < (2+\delta)^2$
$4 - 4\delta + \delta^2 < x^2 <4\delta + \delta^2$.
$-4\delta + \delta^2 < x^2 - 4 <  4 \delta + \delta^2$.
Now as $\delta \le 1$ we know $\delta^2 \le \delta$ and so
$-5\delta < -4\delta + \delta^2 < x^2 - 4 < 4 \delta + \delta^2 < 5\delta$
$-15\delta < 3x^2 - 12 < 15\delta$
$|3x^2 -12| < 15\delta$.
So $\delta \le \frac {\epsilon}{15}$ then 
$|3x^2 - 12| < 15\delta \le 15\frac {\epsilon}{15}=\epsilon$.
And that is the definition of $\lim_{x\to 2} 3x^2 = 2$.
(I do advise rereading the definition and understanding what it means:
(That is for ANY small distance $\epsilon > 0$ no matter how small, we can determine that there is some distance $\delta > 0$ so then whenever we can choose ANY $x$ within $\delta$ of $2$, that is to say, if $|x -2| < \delta$, we will then know that it MUST also be that $3x^2$ is within $\epsilon$ of $12$.  That is $|3x^2 - 12| < \epsilon$.)
A: Let $|x-2|\lt 1$ , then 
$-1<x-2<1$, or $3<x +2<5$.
$|3x^2-12| = 3|x-2||x+2|$.
Let $\epsilon >0$ be given.
Choose $ \delta = \min (1,\epsilon/(15))$.
Then $|x-2| \lt \delta$ implies
$3|x-2||x+2| \lt $
$(3)(5)|x-2| \lt (15)\delta \lt \epsilon$.
A: What you have given is sometimes called a "synthetic" proof- you start with what you want to prove, $|3x^2- 12|< \epsilon$, and arrive at $\gamma\le \frac{\epsilon}{15}$.  Of course, you can't prove something by assuming it is true!  This is, still, a valid proof because every step is reversible.  That is, the "real proof" starts by taking $\gamma$ to be any number less than $\frac{\epsilon}{15}$ then works backwards to $|3x^2- 12|< \epsilon$.
