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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?

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  • $\begingroup$ Choose $\delta=\epsilon/5$ for $\epsilon<1$ $\endgroup$ – Fakemistake Jan 23 '18 at 20:46
  • $\begingroup$ "Am I stating this correctly so far / understanding the delta-epsilon relationship and goals?" For the most part. But there are two considerations you are eliding. i)$\epsilon$ doesn't have to be $.5$ but could be $.05$ or $.000000005$ of $5\times 10^{534}$. It can be any value. ii) And $\delta$ ... well it's best not to think of delta as a constant, but as a number whose value is determined by the value of $\epsilon$. $\endgroup$ – fleablood Jan 23 '18 at 20:58
  • $\begingroup$ If $\epsilon = .5$ then just let $\delta$ be something really really small. If $\delta = .01$ then $|x - 2| < \delta$ means $1.99 < x < 2.01$ so $3.9601< x^2 < 4.0401$ so $-.0399< x^2 - 4 <.0401$ so $|x^2 - 4| < .05 < .5$. So.... that doesn't tell you much about how to do it in general. Does it? $\endgroup$ – fleablood Jan 23 '18 at 21:04
  • $\begingroup$ @fleablood I tried using a graph instead and edited my post. Does this approach make sense? $\endgroup$ – Aruka J Jan 23 '18 at 21:05
  • $\begingroup$ I understand $\epsilon$ can be any positive value, I'm just picking one arbitrarily, fixing some $\epsilon$, and then finding the largest $\delta$ that makes $|f(x)-L| < \epsilon$ hold. We could pick something very small and it would work but that doesn't show how large we could get away with going. $\endgroup$ – Aruka J Jan 23 '18 at 21:05
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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

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

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    $\begingroup$ I think your result is the same as mine correct? $\delta = \sqrt{4.5} - 2$? (I picked $\epsilon = 0.5$) $\endgroup$ – Aruka J Jan 23 '18 at 21:07
  • $\begingroup$ @ArukaJ Yes but you need to prove it in general for genrec $\epsilon$ and $\delta$ values. $\endgroup$ – gimusi Jan 23 '18 at 21:19
  • $\begingroup$ For the third time. !!!STOP!!! picking specific values of $\epsilon$. It will NOT help you. $\endgroup$ – fleablood Jan 23 '18 at 21:32
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    $\begingroup$ @gimusi Oh, yeah. Absolutely. Looks good. i haven't gone through it with a fine tooth comb but, yeah, that's exactly how do it. So barring any arithmetic errors, It's good. My comment was for Aruka who seems hell bent on finding specific deltas for specific elements and not actually trying to find a general formula for deltas for all possible epsilons. $\endgroup$ – fleablood Jan 23 '18 at 21:39
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    $\begingroup$ Well, it's fine to get an insight but there comes a point where you will have to say, insight is done, now we must do a proof. And you will NEVER be allowed to say "It worked for all examples I tried therefore it must be true" as a proof. $\endgroup$ – fleablood Jan 23 '18 at 21:45
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Hint $$|x^{ 2 }-4|=\left| x-2 \right| \left| x-2+4 \right| <{ \left| x-2 \right| }^{ 2 }+4\left| x-2 \right| $$

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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.

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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.

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  • $\begingroup$ Are delta-epsilon proofs more art than science, is that sort of what you are saying? There is no standard methodical way to get the answer? $\endgroup$ – Aruka J Jan 23 '18 at 20:39
  • $\begingroup$ Is bicycle riding an art? No, it is a skill. delta-epsilon proofs are a skill. There are many skills that you can only pick up by actually doing. Swimming, bike-riding are two out of three immediate examples that come to mind. $\endgroup$ – uniquesolution Jan 23 '18 at 20:41
  • $\begingroup$ Proofs in general are like this. There are some approaches that work more often than others, but it's more about the intuition of what works than following a particular step-by-step approach. In this case, the best advice you can get is to poke around with the triangle inequality and when you get something that works, try to understand what made it work. $\endgroup$ – AlexanderJ93 Jan 23 '18 at 20:43

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