# Spivak's Calculus (Chapter 5, Problem 41)

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

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

Then

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

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

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
$|x - a| < \delta\Longrightarrow \sqrt{a^2 - \epsilon} < x < \sqrt{a^2 + \epsilon}$
I placed your equation within \$\$, to denote Latex math. You can view the edit if needed. –  Calvin Lin Jun 19 at 21:12
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 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 at 6:16