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Delta-epsilon proof (book example)

  1. Nov 22, 2008 #1
    [itex]\lim_{x \to 0}[/itex] x^(1/3)

    I know that [itex]\delta = {\epsilon}^3[/itex]

    the book gives an example:

    [itex]\lim_{x \to 2}[/itex] (3x - 2) = 4 and you chose [itex]\delta = \frac{\epsilon}{3}[/itex]
    so

    0 < |x-2| < [itex]\delta = \frac{\epsilon}{3}[/itex]

    implies

    |(3x - 2) - 4| = 3|x-2| < [itex]3 (\frac{\epsilon}{3}) = \epsilon[/itex]

    so i should get something like:

    [itex] | \sqrt[3]{x} - 0 | = | x - 0 | = \epsilon[/itex]

    But I don't see how you make the connection between [itex] | \sqrt[3]{x} |[/itex] and (|x - 0|) < [itex]\delta[/itex] and I don't see how I can reduce [itex]{\epsilon}^3[/itex] to just epsilon using this style of proof.
     
  2. jcsd
  3. Nov 22, 2008 #2
    Think of it this way, start out with[tex]|x^{\frac{1}{3}}|[/tex] and try to get this to be [tex]\leq |x| < \delta[/tex] (we both know the limit is zero ;)). Your choice of [tex]\delta[/tex] seems to be correct. Now given that [tex]|x| < \delta[/tex] what is [tex]|x^{\frac{1}{3}}|[/tex] less than?

    Good luck
     
  4. Nov 22, 2008 #3
    You are suspecting that x1/3 converges to zero as x approaches 0. So how does one show that. First you pick an arbitrary e>0. Then you want to find an d>0 with the following property.
    Whenever you take an x which differs from zero by at most d and plug this x into the function x1/3, you want to get something which differs from the proposed limit, 0, by at most e.

    The question is how to choose that d, right?
    You want |x1/3-0 |<d for all |x-0|<d. We can simplify this to
    |x1/3|<e for all |x|<d, right?
    Is the first equation not equivalent to |x|<e3? So, how small an x do you have to choose to make sure that this is certified. Well, certainly an x with |x|<e3, right? So what should d be then?:smile:
     
  5. Nov 22, 2008 #4
    Thanks.

    I thought we wanted |x^(1/3) - 0 | < e for all | x - 0| < d. I think I see where to go, though.
     
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