Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

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!

share|improve this answer
    
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
add comment

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...

share|improve this answer
add comment

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} $

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.