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

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 $
$(3)(5)|x-2| \lt (15)\delta \lt \epsilon$.
