How to prove this delta-epsilon proof involving $x^2$?

Question

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.

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

2. 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?

Answer 1

We need to show that $\forall \epsilon>0$ $\exists\delta>0$ such that

$$\forall x\neq2 \quad |x-2|<\delta \implies\left|f\left(x\right)-l\right|<\varepsilon$$

that is

$$|x^2-4|<\epsilon\iff-\epsilon<x^2-4<\epsilon\iff4-\epsilon<x^2<4+\epsilon\iff \sqrt{4-\epsilon}<x<\sqrt{4+\epsilon}\iff \sqrt{4-\epsilon}-2<x-2<\sqrt{4+\epsilon}-2\\\iff |x-2|<min\{\sqrt{4+\epsilon}-2,2-\sqrt{4-\epsilon}\}=\sqrt{4+\epsilon}-2=\delta \quad \square$$

Answer 2

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

Answer 3

It is my honest opinion that your question lies in the realm of questions of the type: "what is the best approach for riding a bicycle?" - Most answers you will receive will send you snapshots of happy riders; but you don't really want these answers, do you? I suggest you get on the bike, and keep falling until you don't.

Answer 4

If $|x- 2| < \delta$ then

$- \delta < x -2 < \delta$

$2 - \delta < x < 2 + \delta$. Let's assume for the moment that $\delta < 2$.

$(2- \delta)^2 < x^2 < (2+ \delta)^2$

$4 - 4\delta + \delta^2 < x^2 < 4 + 4\delta + \delta^2$

$-4\delta + \delta^2 < x^2 - 4 < 4\delta + \delta^2$.

$-4 \delta - \delta^2 < x^2 - 4 < 4\delta + \delta^2$

$|x^2 - 4| < |4\delta + \delta^2| = 4\delta + \delta^2$ (because $\delta$ is positive)

So we want $\epsilon \ge 4\delta + \delta^2$. Given that we know what $\epsilon$ is, can we find a way of figuring out $\delta$ in terms of $\epsilon$ so that that would be true?

If we assume $\delta \le 1$ then $\delta^2 \le \delta$ so $5\delta \ge 4\delta + \delta^2$.

So if we choose any $\delta$ so that i) $\delta < 2$ and ii) $\delta \le 1$ and iii) $5\delta < \epsilon$ that will do.

So for any $\delta < \min (\frac \epsilon 5, 1)$ that will do.

So to do the proof:

For any $\epsilon > 0$, let $\delta = \min (\frac \epsilon 5, 1)$

Then if $|x - 2| < \delta$ implies by all the work we did above that

$-5\delta \le -4\delta -\delta^2 < -4\delta + \delta^2 < x^2 -4 < 4\delta + \delta^2 \le 5\delta$ so

$|x^2 - 4| < 5\delta \le \epsilon$.

And that's the proof.