Elementary question regarding an $\epsilon$-$\delta$ proof. It's a stupid question, but the answer eludes me.
Suppose I want to show that 
$$\lim_{x \to 2} (2x^2+3)=11$$
Then I want to show that for every $\varepsilon>0$ there is a $\delta>0$ such that 
$$0<|x-2|<\delta \implies  |2x^2+3-11|<\varepsilon.$$ 
Starting from the last part I write 
$$|2x^2+3-11|=2|x^2-4|=2|x+2||x-2|$$
and 
$$|x+2||x-2|<\frac{\varepsilon}{2}.$$
Now the problem: people say let's consider $\delta=1$, so $|x-2|<\delta$, $1<x<3$ and $3<x+2<5$
and then 
$$|x+2||x-2|<5|x-2|$$ 
and pick 
$$|x-2|<\frac{\varepsilon}{10}$$ 
in this way 
$$\delta=\min(1,\frac{\varepsilon}{10})$$
What I find difficult to understand (even if I see it is a more conservative choice) is why considering 5 and not 3, If 
$$|x+2||x-2|<\frac{\varepsilon}{2}$$ 
shouldn't the correct order be
$$3|x-2|<|x+2||x-2|<\frac{\varepsilon}{2}$$
and so
$$3|x-2|<\frac{\varepsilon}{2} \Rightarrow |x-2|<\frac{\varepsilon}{6}$$
Choosing 5 I substitute a value greater than $|x+2|$ in the expression, and how can I know if the product remains smaller than $\frac{\varepsilon}{2}$? To me it is like writing $10<100$ then $10<200$ and deduce that $200<100$...
I know there must be something wrong in my argument because $\frac{\varepsilon}{10}$ produce a smaller neighbourhood, and furthermore, if I write 
$$10^{-999} |x-2|< 3|x-2|<|x+2||x-2|<\frac{\varepsilon}{2}$$ 
I obtain
$$|x-2|<\frac{10^{999} \varepsilon}{2}$$ 
which is silly because it would imply a huge (ad libitum) value in the $\delta$ and the choice would be always 1 no matter the $\varepsilon$...
 A: Nobody? I'll try to answer my own question then, hoping for some useful comment and/or correction.
Let's start from the end to try finding the right $\delta$ for the chosen $\varepsilon$:
$$|x-2||x+2|<\frac{\varepsilon}{2}$$
so
$$|x-2|<\frac{\varepsilon}{2|x+2|}$$
If I take $|x-2|<1$ (this is the tricky part because it seems to invert the role of $\varepsilon$ and $\delta$) I obtain
$$1<x<3$$
that is (picking 3 on the right side)
$$\frac{\varepsilon}{2|x+2|}>\frac{\varepsilon}{10}$$
Now I have two things:
$$|x-2|<\frac{\varepsilon}{2|x+2|}$$
and (when $|x-2|<1$)
$$\frac{\varepsilon}{10}<\frac{\varepsilon}{2|x+2|}$$
In this way, if $\varepsilon>10$, that is $\frac{\varepsilon}{10}>1$, we have that taking $|x-2|<1$ gives
$$|x-2||x+2|<|x+2|<5<\frac{\varepsilon}{2}$$
and so $\delta=1$ would be a suitable choice. 
Otherwise, if $\varepsilon<10$, that is $\frac{\varepsilon}{10}<1$, taking $\delta=\frac{\varepsilon}{10}$ is good because in this case we know that
$$\frac{\varepsilon}{10}<\frac{\varepsilon}{2|x+2|}$$
and choosing $|x-2|<\delta=\frac{\varepsilon}{10}$ assures that 
$$|x-2|<\frac{\varepsilon}{2|x+2|}$$
