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I've been learning about $\epsilon$-$\delta$ proofs and attempted to come up with my own proof that $$ \lim_{x \to 3} x^2 = 9 $$ exists (I did use some help from some textbooks). Is my proof valid and free of redundancies? My proof:

Scratch work: Written formally, this is: $$ \forall \epsilon > 0, \exists \delta > 0 \text{ s.t. } 0 < |x - 3| < \delta \implies |x^2 - 9| < \epsilon $$ We start off by simplifying the conclusion of the implication: $$ |x^2 - 9| = |x + 3|\cdot |x - 3| $$ Now we have $|x + 3| \cdot |x - 3| < \epsilon$. Because $|x - 3| < \delta$, we have $|x - 3| \cdot |x + 3| < \delta |x + 3|$. In an attempt to find a suitable $\delta$, we let $\delta|x + 3| < \epsilon$. Solving for $\delta$, we get $\delta < \frac{\epsilon}{|x + 3|}$. There is a problem though: $\delta$ is defined in terms of $\epsilon$ and the randomly chose point $x$. The definition of $\delta$ can only depend on $\epsilon$. To get around this problem, we will have to estimate the size of $|x + 3|$.

We start of by assuming that $\delta < 1$, which implies that $|x - 3| < 1$. Solving for $x + 3$, we get $5 < x + 3 < 7$. We now know that $|x + 3| < 7$ when $\delta < 1$. Since $|x^2 - 9| < \delta|x + 3|$ and $|x + 3| < 7$ (if $\delta < 1$), we have $|x^2 - 9| < 7 \delta$. In an attempt to find a suitable $\delta$, we let $7 \delta < \epsilon$. Solving for $\delta$, we get $\delta < \frac{\epsilon}{7}$ when $\delta < 1$. So we have now deduced two restrictions: $\delta < 1$ and $\delta < \frac{\epsilon}{7}$. To satisfy both restrictions, we let $\delta < \min \left\{ 1, \frac{\epsilon}{7} \right\}$. When $\epsilon > 7$, then $\delta < 1$ and when $\epsilon < 7$, we have $\delta < \frac{\epsilon}{7}$. We can now write up the proof.

Proof: For every $\epsilon > 0$ there exists a $0 < \delta < \min \left\{ 1, \frac{\epsilon}{7} \right\}$ such that $0 < |x - 3| < \delta \implies |x^2 - 9| < \epsilon$. There are two things that can happen: $\delta < 1$ and $\delta < \frac{\epsilon}{7}$.

Case 1 - $\delta < 1$: The implication's hypothesis is $0 < |x - 3| < \delta$. Multiply both sides by $|x + 3|$ to get $0 < |x^2 - 9| < |x + 3| \delta$. Because $\delta < 1$, we know (from our scratchwork) that $|x + 3| < 7$ and $\delta$ is also less that $\frac{\epsilon}{7}$. This means that $0 < |x^2 - 9| < |x + 3| \delta < 7 \delta < [7 \frac{\epsilon}{7} = \epsilon]$.

Case 2 - $\delta < \frac{\epsilon}{7}$: For $\delta$ to be less that $\frac{\epsilon}{7}$, we must have that $\frac{\epsilon}{7} < 1$ (as per the definition of $\min$). Because $\delta < \frac{\epsilon}{7}$ and $\frac{\epsilon}{7} < 1$ we have $\delta < 1$. Look to Case 1 for what follows after.

$\square{}$

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    $\begingroup$ I believe there is a typo in your first displayed equation: you wrote $$\lim_{\color{red}{c\to3}} x^2 = 9$$ $\endgroup$
    – Joe
    Jul 26, 2021 at 21:55
  • $\begingroup$ Is there a question here? $\endgroup$ Jul 26, 2021 at 21:56
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    $\begingroup$ I think you have a very good handle on this! In the "sketch work" when you wrote "Now we have |x+3|⋅|x−3|<ϵ. Because |x−3|<δ, we..." I was sure where you were coming from our going to as we didn't have anything yet, but it became clear as I read what you were doing (attempting to find nesc and/or restrictions on $\delta$). In any event I very much like that after doing all the sketch work you presented a formal and correct proof frome the start. $\endgroup$
    – fleablood
    Jul 26, 2021 at 22:52
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    $\begingroup$ Oh.... An improvement would be to combine and not consider cases. Just note: If $|x-3|<\delta < 1$ then $0 < |x+3| < 7$ so $|x^2-9|=|x-3|\cdot |x+3| < \delta\cdot 7 < \frac \epsilon 7\cdot 7 = \epsilon$. That's all you have to say.... After all you don't have two cases where one or the other is true. You have one case where both are true. $\endgroup$
    – fleablood
    Jul 26, 2021 at 22:57
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    $\begingroup$ ... Just to be perverse. If $\delta < 1 < \frac \epsilon 7$ then $|x^2-9|=|x-3||x+3| < \delta \cdot 7 < 1\cdot 7 =7$. But as $ \frac{\epsilon}7 > 1$ we know $\epsilon > 7$ so $|x^2-9| < 7 < \epsilon$..... But it is perverse to bother trying to consider cases where $\epsilon$ is large. $\endgroup$
    – fleablood
    Jul 26, 2021 at 23:09

3 Answers 3

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Congratulations, your proof is valid! Here are a few minor improvements you could make:

  • It is simpler to set $\delta=\min\left\{1,\frac{\varepsilon}{7}\right\}$ than to require that $\delta<\min\left\{1,\frac{\varepsilon}{7}\right\}$.
  • If $\delta=\min\left\{1,\frac{\varepsilon}{7}\right\}$, then it is automatically true that $|x-3|<1$ and $|x-3|<\frac{\varepsilon}{7}$. Therefore, there is no need to split your proof into cases. You can write it up as follows:

If $\delta=\min\left\{1,\frac{\varepsilon}{7}\right\}$, then $|x-3|<1$ and so $|x+3|<7$. Since $|x-3|<\frac{\varepsilon}{7}$, $$ |x^2-9|=|x-3||x+3|<\frac{\varepsilon}{7} \cdot 7=\varepsilon \, .$$

  • The final tip is by far the most important. Please use \varepsilon instead of \epsilon: it will make the users at TeX.SE eternally grateful.

Exercise: can you generalise your proof to show that the function $x\mapsto x^2$ is continuous, i.e. for every real number $a$, $\lim_{x \to a}x^2=a^2$? The basic idea is the same: start by requiring that $|x-a|<1$ and use this to come up with an upper bound for $|x+a|$. Note that while $\delta$ cannot depend on $x$, it can depend on $a$.

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  • $\begingroup$ I have two questions: (1) Why can we let $\delta = \min\{1, \frac{\varepsilon}{7}\}$ (2) How does $\delta = \min\{1, \frac{\varepsilon}{7}\}$ imply that $|x - 3| < 1$ and $|x - 3| < \frac{\varepsilon}{7}$ both are true? $\endgroup$
    – obr
    Jul 26, 2021 at 22:19
  • $\begingroup$ @obr: I'm sorry but I don't understand your first question. Is it possible if you elaborate please? We need to show that there is a $\delta>0$ such that if $|x-3|<\delta$, then $|x^2-3^2|<\varepsilon$. One value of $\delta$ that works is $\min\left(1,\frac{\varepsilon}{7}\right)$, and we know that it works because of the proof. Of course, you can pick a $\delta$ that is even smaller than $\min\left(1,\frac{\varepsilon}{7}\right)$, but you don't need to. $\endgroup$
    – Joe
    Jul 26, 2021 at 22:27
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    $\begingroup$ @obr: Regarding your second question, let $x$ be the smaller number out of $1$ and $\frac{\varepsilon}{7}$, and $y$ be the larger of the two. Since $|x-3|<\min\left(1,\frac{\varepsilon}{7}\right)$, this means that $|x-3|<x$. And since $x\le y$, we get that $|x-3|<x\le y$. Hence, it is also true that $|x-3|<y$. Does that help? $\endgroup$
    – Joe
    Jul 26, 2021 at 22:28
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    $\begingroup$ @obr Why wouldn't you be able to let $\delta = \min\{1, \frac{\varepsilon}{7}\}$? It's a strictly positive number, and as Joe has shown, it works to prove the limit, which are the only two restrictions on what $\delta$ can be. $\endgroup$ Jul 26, 2021 at 22:47
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There is no need to consider several cases. Define $\delta=\min\left\{\frac\varepsilon7,1\right\}$. Then, if $|x-3|<\delta$, it follows from your computations that$$\left|x^2-9\right|=|x-3||x+3|<\frac\varepsilon7\times7=\varepsilon.$$

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The scratch work looks good, but in the final proof there is no need to split into cases. You can just write

Given $\epsilon>0,$ take $\delta$ such that $0<\delta<\min\{1,\frac\epsilon7\}.$ Then, when $|x-3|<\delta,$ one has $$ |x^2-9| = |x+3||x-3| = |(x-3)+6| |x-3| < (\delta+6)\delta < (1+6)\frac\epsilon7=\epsilon. $$

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