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Fallacious Limit Proof

  1. Jul 12, 2010 #1

    jgens

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    1. The problem statement, all variables and given/known data

    Consider the following proof: Suppose that we want to prove that [itex]\lim_{x \to a} x^2 = a^2[/itex] for [itex]a > 0[/itex]. Given [itex]\varepsilon > 0[/itex], we simply let [itex]\delta[/itex] be the minimum of [itex]\sqrt{a^2 + \varepsilon} - a[/itex] and [itex]a - \sqrt{a^2 - \varepsilon}[/itex]; then [itex]0 < |x-a| < \delta[/itex] implies that [itex]\sqrt{a^2 - \varepsilon} < x < \sqrt{a^2 + \varepsilon}[/itex], so [itex]a^2 - \varepsilon < x^2 < a^2 + \varepsilon[/itex], or [itex]|x^2 - a^2| < \varepsilon[/itex].

    This "proof" is entirely fallacious. Wherein lies the fallacy?

    2. Relevant equations

    Definition of a limit.

    3. The attempt at a solution

    I haven't really been able to come up with any reason why the proof is fallacious. Sure, his choice of [itex]\delta[/itex] implies that [itex]\varepsilon < a^2[/itex] which contradicts the "for every [itex]\varepsilon > 0[/itex]" clause in the definition of a limit; however, it seems that this could be dealt with by the following argument: If there is in fact a [itex]\delta[/itex] which works for [itex]\varepsilon < a^2[/itex], then it would follow that this same [itex]\delta[/itex] would work for [itex]\varepsilon \geq a^2[/itex] as well. So, I'm stuck.

    I would appreciate it if anyone could point me in the right direction. Thanks!
     
  2. jcsd
  3. Jul 12, 2010 #2
    I noticed that your [tex]\delta[/tex] dosn't depend on [tex]a[/tex] at all.
    because [tex]a - \sqrt{a^2}=0[/tex] and in the same manner you can write [tex]\delta[/tex] from [tex]y-y - \varepsilon[/tex] to [tex]y-y + \varepsilon[/tex]
    and prove the same result for any variable y no matter its connection.

    just my thought :)
     
    Last edited: Jul 12, 2010
  4. Jul 12, 2010 #3

    jgens

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    Sorry, the LaTeX code didn't display things right. Hopefully this works better . . .

    [tex]\delta = \min{\sqrt{a^2 + \varepsilon} - a, a - \sqrt{a^2 - \varepsilon}}[/tex]
     
  5. Jul 12, 2010 #4

    hunt_mat

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    My guess would be the following (it's been a while since I did this) is that the proof only needs to be for a>0. So
    [tex]
    |x^{2}-a^{2}|=|(x-a)(x+a)|<\delta |x-a+2a|<\delta (\delta +2a)
    [/tex]
    So choose your epsilon accordingly so that [tex]\delta (\delta +2a)<\varepsilon[/tex]. I don't think that your delta should be a function of epsilon, it should be the other way around.
     
  6. Jul 12, 2010 #5
    How is the square root defined (is it a function) and how does it exist
     
    Last edited: Jul 12, 2010
  7. Jul 12, 2010 #6

    jgens

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    hunt_mat, I already know how to correctly prove that [itex]\lim_{x \to a} x^2 = a^2[/itex]. I need figure out why the "proof" given by the author is fallacious. And, at least from the calculus book that I'm working out of, [itex]\delta[/itex] is usually expressed as a function of [itex]\varepsilon[/itex], so I don't think that that's the problem.
     
  8. Jul 12, 2010 #7

    jgens

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    The author hasn't really defined the square root function yet since we have yet to prove that every non-negative real number has a square root. However, he operates on the assumption that it is a well-defined function until he introduces the least upper bound property several chapters after this one.
     
  9. Jul 12, 2010 #8
    Just think about how the square root function would be defined in relation to the function x^2. It can be related to the intermediate value theorem and some other properties of the inverses of functions.
     
  10. Jul 12, 2010 #9

    jgens

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    I'm not really sure where you're going with this. Sure, I know that if [itex]\sqrt{x}[/itex] has a real out put, then [itex]x \geq 0[/itex]. I know that if [itex]x < y[/itex], then [itex]\sqrt{x} < \sqrt{y}[/itex]. I'm sorry, I don't understand what you're asking me to do.

    As an aside, while I'm familiar with the IVT and inverse functions, the author of this textbook doesn't do much to formally develop them until two or three chapters after this one. So, I don't think that the answer should involve anything which deals extensively with them.
     
  11. Jul 12, 2010 #10

    Office_Shredder

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    I'm a bit confused... is the latex error of the square root not holding the epsilon throughout the entire proof or only in the definition of delta?
     
  12. Jul 12, 2010 #11
    Blargh I remember this question from Spivak, and the intended answer is in fact what nike5 first suggested. I believe Spivak proves the existence of square roots by using the intermediate value theorem applied to f(x) = x^2. Of course the key hypothesis in the theorem is the continuity of the function, which is basically what you're trying to prove. Hence the element of circularity is the "fallacy".

    But the existence of square roots can be proved directly from the least upper bound axiom, so the proof is only fallacious within the context of Spivak's presentation of the material. For instance, Rudin proves the existence of positive roots a few pages into his text. So really, there was nothing you could have done to answer this unless you had anticipated that Spivak wasn't going to prove the existence of square roots from first principles, and such an anticipation is kind of silly.
     
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