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Proving x^(1/3) is continuous.

  1. Oct 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove x^(1/3) is continuous on all of ℝ.

    3. The attempt at a solution

    I've essentially gotten everything to the following point:

    [|x-c|/|x2/3 + (cx)1/3 + c2/3|]<ε

    I'm having trouble coming up with a lower bound for the denominator. Any help?

    Thanks in advance!
     
  2. jcsd
  3. Oct 24, 2012 #2

    Zondrina

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    Did it ask you to use the epsilon delta definition? Really?

    Why don't you use the definition of the derivative to show that x1/3 is continuous everywhere?

    First off, notice that x1/3 is continuous [itex]\forall x \in ℝ[/itex]

    Now you want to show that its derivative, [itex]\frac{1}{3} x^{- \frac{2}{3}}[/itex] is also continuous [itex]\forall x \in ℝ[/itex].

    Notice though, if you pick x = x0 and apply the definition of the derivative to x1/3, you will find that x1/3 has continuous derivatives everywhere except possibly the origin?

    Now what happens when you apply the definition of the derivative on x1/3 for x = 0? Does your limit exist?
     
  4. Oct 24, 2012 #3
    Thanks for your quick response! Unfortunately, we haven't gotten to the chapter on derivatives yet, and we're only allowed to use an epsilon-delta proof..
     
  5. Oct 24, 2012 #4

    Zondrina

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    So then really what you want to show is :

    [itex]\lim_{x→a} x^{\frac{1}{3}} = a^{\frac{1}{3}}[/itex]

    So apply your definition :

    [itex]\forall ε > 0, \exists δ > 0 \space | \space 0 < |x-a| < δ \Rightarrow |x^{\frac{1}{3}} - a^{\frac{1}{3}}| < ε[/itex]

    Now start by massaging [itex]|x^{\frac{1}{3}} - a^{\frac{1}{3}}|[/itex] into [itex]|x-a| < δ[/itex]
     
  6. Oct 24, 2012 #5
    Sorry, I should've showed my earlier steps. I already started working on [itex]|x^{\frac{1}{3}} - c^{\frac{1}{3}}|[/itex].

    From there, I multiplied the numerator and denominator to get to [|x-c|/|x2/3 + (cx)1/3 + c2/3|]<ε

    Now I have my |x-c|, but I can't come up with a lower bound on the denominator. I've tried a bit of algebraic manipulation, but to no avail.
     
  7. Oct 24, 2012 #6

    Zondrina

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    Ah I see, so. After you multiply by the conjugate you know that |x-c| < δ right? So that mess [|x-c|/|x2/3 + (cx)1/3 + c2/3|] < δ/|x2/3 + (cx)1/3 + c2/3|

    ( I'm assuming you've done your arithmetic correctly here, don't have time to check at the moment ).

    Now you want to ask yourself, how big does :

    |x2/3 + (cx)1/3 + c2/3| get?

    Apply your triangle equality now.
     
  8. Oct 24, 2012 #7
    I'm actually trying to see how small it gets. Hence, I need |x2/3 + (cx)1/3 + c2/3| to be greater than some fixed value.
     
  9. Oct 24, 2012 #8
    Just so you know. It is ok to choose a two values of x near c to find this upper limit. After all, it is the only place of interest.
     
  10. Oct 24, 2012 #9
    Do you mean going ahead and saying δ<1 (or some number) so that |x-c|<1, implying x>(1-c)?

    I tried something of that sort but ran into some issues, because the bound still depended on the value of c.
     
  11. Oct 24, 2012 #10
    Yes.

    Let sigma be less than 1. The |x-c|<1.
    Thus, c-1<x<1+c. So substitute for which ever x to minimize
    |x^(2/3) + (cx)^(1/3) + c^(2/3)|.

    Clearly, for c-1 substituted for x is a lower bound of that expression.

    Continue from there and also note that I'm assuming the expression is correct.
     
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