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In Chapter 5, Problem 41, Spivak provides an alternative way to prove that

$$\lim_{x \rightarrow a} x^2 = a^2\,\,,\,\,a > 0$$

Given $\,\epsilon > 0\,$ let

$$\delta = \min\left\{\sqrt{a^2 + \epsilon} - a, a - \sqrt{a^2 - \epsilon}\right\}$$


$$|x - a| < \delta\Longrightarrow \sqrt{a^2 - \epsilon} < x < \sqrt{a^2 + \epsilon}\Longrightarrow a^2 - \epsilon < x^2 < a^2 + \epsilon\,\,,\, |x^2 - a^2| < \epsilon$$

Then he goes on to claim that this proof is fallacious. But wherein lies the fallacy?

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I don't see any fallacy. – Flanders Sep 14 '12 at 14:03
Spivak's Calculus1994 page 100 – noname1014 Sep 14 '12 at 14:07

In Spivak's book, this limit fact (later stated as: function $x^2$ is continuous) is proved quite early. Before the existence of square-roots is known. Indeed, continuity of the function $x^2$ will later be used to prove existence of square-roots. So an argument with square-roots here would be circular reasoning!

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circular reasoning is very hard to find. – noname1014 Sep 14 '12 at 14:25
I agree: if you do not know that $x \mapsto x^2$ is increasing and continuous, it is rather hard to define $\sqrt{x}$. – Siminore Sep 14 '12 at 14:35
I think this is a fake "fallacy". See new answer. – zyx Jun 3 '14 at 19:44

It is wrong to call the proof "fallacious". The comments in the solution book,

"41. How do we know that $\sqrt{a^2 - \epsilon}$ and $\sqrt{a^2 + \epsilon}$ exist? In Chapter 7 we prove (Theorem 8) that every positive number has a square root, but the proof of this theorem uses the fact that $f(x)=x^2$ is continuous, which is essentially what we are trying to prove. In fact the existence of square roots is essentially equivalent to the continuity of $f$ --- compare Problem 8-8"

contain several errors.

  • The proof and calculations are correct; the stated value of $\delta$ exists and suffices for the $\epsilon$ argument.

  • The supposed problem with assuming the existence of the square roots does not derive from the proof, but from the sequence chosen for the material in the book. The quantities with square roots do exist as real numbers, and using those values for $\delta$ makes the proof work.

  • The existence of square roots does not logically depend on proving continuity of $f(x)=x^2$. Other books give the existence statement as an exercise on least upper bounds (of the positive rational solutions of $x^2 < a$) or as convergence of the Babylonian square root iteration.

  • In epsilon-delta proofs, including the one Spivak wrote down for this exercise, equations like $\delta = \min(A,B,C,\dots, K)$ do not necessarily assume that the bounds $A,B,C...$ all exist. It is only required that all inequalities $\delta \leq A$ and $\delta \leq B$ can be satisfied at once. Some of the upper bounds might not exist for all values of $a$ and $\epsilon$ (in this problem, take $\epsilon > a^2$ and one of the square roots becomes imaginary).

  • Completing the proof without taking existence of roots as an assumption, does not require an existence proof for the root. All we need is solutions to conditions like $\delta \leq \sqrt{a^2+\epsilon} - a$, which can be done by squaring,

$(\delta + a)^2 < a^2 + \epsilon$

an inequality that is easy to achieve with pure algebraic epsilon-delta calculations. No properties of real numbers such as convergence or completeness are necessary.

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I agree that this seems a bit unfair, and in my opinion something like this should have been in textual material. I edited something near the end of your question, with the intent to make one of your points more explict. – Dave L. Renfro Jun 3 '14 at 19:48
Thanks. I un-did the edit, in that I really did mean to write the conditions that should ultimately be satisfied as $\delta \leq A$, etc in their original form (including square roots or other possibly undefined quantites) and then show that on a nonempty interval of values of $\delta$, those conditions (and/or whichever inequalities they are used to derive) are satisfied. – zyx Jun 3 '14 at 19:59
I disagree with you. Just because someone else may know a proof that $\sqrt x$ exists for all positive $x$ does not mean that it is a valid proof in the context of the book itself. I can certainly pull all sorts of mathematical knowledge out of some book, but that does not make it a valid proof in the context of the book/course I am following in a linear fashion. – Ted Shifrin Jun 3 '14 at 20:16
@TedShifrin, was this one of the problems you contributed to the later editions of the book? – zyx Jun 8 '14 at 5:07
Nope, I'm not to blame for this one. It's all Mike Spivak. :) – Ted Shifrin Jun 8 '14 at 11:27

This may just be tom - foolery that I'm pointing out, but if $\epsilon > a^2$ then I'm not too sure what $\delta$ you would pick...

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I think the problem is that you can't say which of the two values of $\delta$ is the minimum so you can´t assume

$ |x - a| < \delta\Longrightarrow \sqrt{a^2 - \epsilon} < x < \sqrt{a^2 + \epsilon} $

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I placed your equation within \$ \$, to denote Latex math. You can view the edit if needed. – Calvin Lin Jun 19 '13 at 21:12
Even if we couldn't say which of the two expressions is the smaller (in fact, we can, but never mind), how would that prevent us from deducing the implication that you say "you can't assume"? – Andreas Blass Jun 19 '13 at 21:40
i am sorry my english is not as good as i wish. Could you explain please how $|x - a| < \delta\Longrightarrow \sqrt{a^2 - \epsilon} < x < \sqrt{a^2 + \epsilon}$ – user83140 Jun 19 '13 at 21:57
@user83140: First use the definition of absolute value to show that $|x-a|<\delta$ is equivalent to $a-\delta<x<a+\delta$. Then apply the definition of $\delta$. – Andreas Blass Jun 28 '13 at 6:16
@AndreasBlass if $ \sqrt{a^2+\epsilon}-a < a-\sqrt{a^2-\epsilon}$ then $\delta=\sqrt{a^2+\epsilon}-a$. So, if $0 < |x-a|<\sqrt{a^2+\epsilon}-a\Longrightarrow a-\sqrt{a^2+\epsilon}-a<x<\sqrt{a^2+\epsilon}+2a$. From here, how can you conclude that $|x^2-a^2|<\epsilon$? Same occurs on the other case. – Allonsy Oct 16 '15 at 15:34

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