How to prove this delta-epsilon proof involving $x^2$? Unlike my last question I want to try something where $x$ can be any real value in $f(x)$ so it's not just $x \geq 0$. I want to fix an $\epsilon$ and find the largest $\delta$ I can get away with using to make the necessary inequalities hold.
$$\lim_{x \rightarrow 2} x^2= 4$$
Say I pick some $\epsilon = 0.5$ which means I need to find some $\delta > 0$ such that for all $x$ satisfying $0 < |x-2| < \delta$, the inequality $|f(x) - L| = |x^2 - 4| < 0.5$ holds.


*

*Am I stating this correctly so far / understanding the delta-epsilon relationship and goals?

*How do I pick the right $\delta$ for something like this?
Trying to simplify:
$|x^2 - 4| < 0.5$
$|x+2||x-2| < 0.5$
$|x-2| < \frac{0.5}{|x+2|}$
Now I'm stuck. Is there a better way to approach these problems? So far I've been trying to manipulate the epsilon inequality so it looks more like the delta one and then try to set the delta and epsilon expressions equal to each other, but maybe there is a more reliable way to prove these relationships?
Update:
Trying another way:
$|x^2 - 4| < 0.5$ simplifies to
$\sqrt{3.5} < x < \sqrt{4.5}$
This gives me two $x$-values away from $a=2$, either $2 - \sqrt{3.5} = .1291...$ or $\sqrt{4.5} - 2 = .1213...$
So if I pick the smallest of the two, $\delta = \sqrt{4.5} - 2$ which satisfies the epsilon condition?

Hint
$$|x^{ 2 }-4|=\left| x-2 \right| \left| x-2+4 \right| <{ \left| x-2 \right|  }^{ 2 }+4\left| x-2 \right| $$
