1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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]

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


    |(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

    I thought we wanted |x^(1/3) - 0 | < e for all | x - 0| < d. I think I see where to go, though.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook