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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
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3 Answers

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
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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
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