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I am currently studying the formal defenition of the limit. One of the examples given by my book is the following: Prove that: $$ \lim_{x \to 3} x^2 = 9 $$ So, using only the defenition of the limit, I have to prove that for every $\epsilon > 0$ there is a $\delta > 0$ for which the following is true: $$ 0 < |x - 3| < \delta \to |x^2 - 9| < \epsilon $$ So after some puzzeling I came up with: $\delta = \frac{\epsilon}{|x + 3|}$. And I though this was correct: under the assumption that the antecedent is true, we can make the following construct: $$ 0 < |x - 3| < \delta \Rightarrow 0 < |x - 3| < \frac{\epsilon}{|x + 3|} \Rightarrow 0 < |x + 3||x - 3| < \epsilon \Rightarrow |x^2 - 9| < \epsilon $$

Q.E.D, I thought. But the book came up with the following solution: $$ \delta = \min{(1,\frac{\epsilon}{7})} $$ Which is a correct solution. So, is mine wrong? Or did the book just provide a different proof?

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  • $\begingroup$ You cannot make $\delta$ depend on $x$. $\delta$ tells $x$ how far away from $3$ it can be, and therefore needs to absolutely independent of $x$. Read the sentence "There is a $\delta > 0$ such that for any $x$ such that $|x - 3| < \delta$..." carefuly, and you see that this is true. $\endgroup$ – Arthur Jul 12 '15 at 18:16
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    $\begingroup$ To complement the answers, note that using $1$ is an arbitrary choice. You could just as well chose $42$ and set $\delta = \min \{42, \epsilon/ 48\}$ $\endgroup$ – quid Jul 12 '15 at 18:22
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Your $\delta$ can't depend on $x$, because $x$ is varying, and so is your $\delta$, but $\delta$ was supposed to be constant. That's the mistake. Let's do a reverse engineering, supposing that $\delta < 1$ (this usually helps).

If $|x-3|<\delta$, then $|x-3|<1 \implies |x| < 3+1 = 4$. And: $$|x^2-9|= |x+3||x-3|\leq (|x|+3)\delta < 7\delta,$$so that $\delta = \min (1, \epsilon/7)$ will work. So far, just ideas.

Now we check: let $\epsilon > 0$, choose $\delta = \min (1,\epsilon/7)>0$ and take $x \in \Bbb R$ such that $|x-3|<\delta$. Since $\delta < 1$, we also have $|x|<4$. And in these conditions: $$|x^2-9| = |x+3||x-3| \leq (|x|+3)\delta < (4+3)\delta = 7\delta < 7 \frac \epsilon 7 = \epsilon.$$

(You can read more about the idea of supposing that $\delta < 1$ in my answer here.)

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First, let's choose $\delta\le1$. Then, we have

$$0<|x-3|<\delta\implies 2<x<4\implies |x+3|<7$$

So, for that choice of $\delta$, we have

$$|x+3|\,|x-3|<7\,|x-3|$$

Now, given $\epsilon>0$, if $|x-3|<\frac{\epsilon}{7}$, then

$$|x^2-9|<\frac{\epsilon}{7}$$

So, whenever $0<|x-3|<\delta=\min(1, \epsilon/7)$ we have

$$|x^2-9|<\frac{\epsilon}{7}$$


NOTE:

There was nothing special about our initial restriction on $\delta$, namely $\delta = 1$. We could have initially chosen any number, say $\delta = 1/2$, or $\delta = 10^{-6}$. So, let's first choose $0<\delta = \delta'$.

Following the previous development we find that for a given $\epsilon >0$,

$$|x^2-9|<\epsilon$$

whenever $|x-3|<\delta =\min\left(\delta',\frac{\epsilon}{6+\delta'}\right)$.


EXAMPLES:

Let's see a couple of examples.

Example 1: For $\delta'=1/2$ we would have for any given $\epsilon$

$$|x^2-9|<\epsilon$$

whenever $|x-a|<\delta=\min\left(\frac{1}{2},\frac{1}{6.5}\epsilon\right)$

Example 2:

For $\delta'=10^{-6}$ we would have for any given $\epsilon$

$$|x^2-9|<\epsilon$$

whenever $|x-a|<\delta=\min\left(10^{-6},\frac{1}{6.000001}\epsilon\right)$

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  • $\begingroup$ Why the down vote????? $\endgroup$ – Mark Viola Jul 13 '15 at 3:42
  • $\begingroup$ Allthough you gave a good proof, you did not answer the question. The question was about wether or not my proof was correct, your answer only proves that the answer given by the book is correct. It says nothing about my proof. I do really appriciate the extended explaination, but I also understand why it is not upvoted. $\endgroup$ – Cheiron Jul 13 '15 at 20:31
  • $\begingroup$ @Cheiron Actually, two of the questions that you asked were, (i) "Which is a correct solution," and (ii) 0r did the book just provide a different proof? So I really thought that my answer was appropriate and useful here to show both a correct solution and the specific solution from your text. For the other question, "So, is mine wrong?," that had already been addressed in another post. But I added another section that shows some examples. I really hope that you find this useful. Let me know how I can further improve my answer. I want to give you the best answer I can. Fair? $\endgroup$ – Mark Viola Jul 13 '15 at 20:43
  • $\begingroup$ It was fair to me a long time ago. At the time I could prove that the book was correct (allthough your explaination deepend my understanding of why the book was correct and gave me a systematic way to seek this kind of solutions), but I could not figure out wether or not my proof was correct. So I made the question more about my proof then about the books' answer, I merely provided that answer to give some additional context. Which I thought was necessary based on some negative experiences in the past. Turns out you guys gave amazing answers I did not even know I was looking for. $\endgroup$ – Cheiron Jul 13 '15 at 21:05

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