# $\epsilon$-$\delta$ proof of $\lim_{x \to 3} x^2 = 9$

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.

$$\square{}$$

• I believe there is a typo in your first displayed equation: you wrote $$\lim_{\color{red}{c\to3}} x^2 = 9$$
– Joe
Jul 26, 2021 at 21:55
• Is there a question here? Jul 26, 2021 at 21:56
• 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. Jul 26, 2021 at 22:52
• 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. Jul 26, 2021 at 22:57
• ... 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. Jul 26, 2021 at 23:09

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 $$a$$.

• 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?
– obr
Jul 26, 2021 at 22:19
• @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.
– Joe
Jul 26, 2021 at 22:27
• @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?
– Joe
Jul 26, 2021 at 22:28
• @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. Jul 26, 2021 at 22:47

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$$\left|x^2-9\right|=|x-3||x+3|<\frac\varepsilon7\times7=\varepsilon.$$

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 $$|x^2-9| = |x+3||x-3| = |(x-3)+6| |x-3| < (\delta+6)\delta < (1+6)\frac\epsilon7=\epsilon.$$