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 $\lim_{x->2}x^{2}+4x=12$
 A: 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 $x=a$.
A: It's correct when you write $|(x^2+4x-12)| = |x+6||x-2|$, but after that, you are not doing the right thing. 
The mistake is in the line : $0 < |x-2| < \delta \implies |x+6| < \epsilon$. You can check that no $\delta$ will satisfy this, the limit does not hold. There also seems to be a confusion between usage of theorem and proceeding by definition. 
You know that $\lim f(x)g(x) = \lim f(x) \lim g(x)$. 
At this point, one crucial thing : if you are using this identity, then there is no point of starting with "let $\epsilon > 0$". What you have to do now, is to show individually that $\lim f(x) \lim g(x) = 0$ (since $f(x)g(x)= x^2+4x-12 \to 0$ is expected as $x \to 2$).
Hence, if you can show that both limits exist, and their product is zero, then you are done.
Now, begin your $\epsilon-\delta$ argument. Let $\epsilon > 0$ (I'll prove both limits together).
Let $\delta = \epsilon$. Then, note that $|x-2| < \delta \implies |(x-2) - 0| < \epsilon$. Hence, $\lim_{x \to 2} (x-2) = 0$.
Let $\delta = \epsilon$. Note that $|x-2|  < \delta \implies |(x + 6) - 8| < \epsilon$. From here, it follows that $\lim_{x \to 2} x + 6 = 8$.
Now, from this you get the product of the limits as $0 \times 8 = 0$. Hence, $\lim_{x \to 2} x^2+4x- 12 = 0$, from which you see that $\lim_{x \to 2} x^2+4x = 12$.
The point here is, first it looked like you started by  definition, and then it looked like you wanted to use a theorem. So if you make your mind up clearly on what you want to do, it would have been easier (like this).
