Prove that $\lim_{x \to2 }x^{2}+4x=12$ Please check my prove
we begin by 
$$x^{2}+4x-12<\epsilon $$
$$|x+6||x-2|<\epsilon $$
I will prove by use property 
$\lim_{x->n}f(x)g(x)=\lim_{x->n}f(x)\lim_{x->n}g(x)$
then we divide into to case
$|x+6|<\sqrt{\epsilon }$ and $ |x-2|<\sqrt{\epsilon }$
and let$\epsilon ,\delta _{1}\delta _{2}>0$
$$0<|x-2|<\delta _{1}\rightarrow |x+6|<\sqrt{\epsilon }$$
Choose $\delta _{1}=\sqrt{\epsilon }$
then $|x+6|<\sqrt{\epsilon }$
$$0<|x-2|<\delta _{2}\rightarrow |x-2|<\sqrt{\epsilon }$$
Choose $\delta _{1}=\sqrt{\epsilon }$
then $|x+6|<\sqrt{\epsilon }$
then use property $\lim_{x->n}f(x)g(x)=\lim_{x->n}f(x)\lim_{x->n}g(x)$
we get
$$|x+6||x-2|<\sqrt{\epsilon }\sqrt{\epsilon }$$
then $\lim_{x->2}x^{2}+4x=12$

Let $\epsilon>0$. 
We wish to find a $\delta>0$ such that if  $\vert x-2\vert<\delta$, then $\vert x^2+4x-12\vert<\epsilon$.
We see that the $x^2-4x-12=(x-2)(x+6)$. Since $x\to2$ we can assume that eventually it will be true that $\vert x-2\vert<1$ which means that
\begin{eqnarray}
-1&<&x-2<1\\
1&<&x<3\\
7&<&x+6<9
\end{eqnarray}
so we see that if $\vert x-2\vert<1$ then $\vert x+6\vert\le9$
So we let $\delta=\min\left\{1,\dfrac{\epsilon}{9}\right\}$
Now if $\vert x-2\vert<\delta$ it follows that
\begin{eqnarray}
\vert x^2+4x-12\vert &=& \vert x+6\vert\cdot\vert x-2\vert\\
&<&9\cdot\frac{\epsilon}{9}\\
&<&\epsilon
\end{eqnarray}
Note that for many limits (there are exceptions) you can find an upper bound $B$ so that 
$$\left\lvert\dfrac{f(x)-L}{x-a}\right\lvert\le B$$
whenever $\vert x-a\vert<1$. In such a case you can let $\delta=\min\left\{1,\dfrac{\epsilon}{B}\right\}$.
This will not work in cases where there is a vertical tangent to $y=f(x)$ at $x=a$.
