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Homework Help: Help on epsilon delta proof of discontinuity

  1. Jul 13, 2011 #1
    1. The problem statement, all variables and given/known data
    Prove the function f(x)= { 4 if x=0; x^2 otherwise
    is discontinuous at 0 using epsilon delta.

    2. Relevant equations
    definiton of discontinuity in this case:
    there exists an e>0 such that for all d>0 if |x-0|<d, |x^2-4|>e

    3. The attempt at a solution
    I'm confused because if we include ALL delta >0, eventually (namely, x around +/- 2) |x^2-4| will be less than e for all e>0 which seems to me to point to it being continuous at 0 when it clearly is NOT. Is there something built into the definition that ignores large values of delta which makes the interval around x too large?

    So my way:
    Assume it is continuous at 0. Let e=1
    Then |x^-4|<1 when |x|<d for some d.

    From here I want to say d<1 and so |x^2-4|>2 for all |x|<d which would be the contradiction but I don't think that's how I formally say it...
  2. jcsd
  3. Jul 13, 2011 #2
    Hi dillingertaco! :smile:

    Your definition of discontinuity is wrong. Continuity in 0 says that

    "For all [itex]\varepsilon >0[/itex], there is a [itex]\delta>0[/itex], such that for all x holds that [itex]|x-0|<\delta~\Rightarrow~|x^2-4|<\varepsilon[/itex]"

    So, the converse of this statement is

    "There is a [itex]\varepsilon >0[/itex], such that for all [itex]\delta >0[/itex], there is an x such that [itex]|x-0|<\delta[/itex] but [itex]|x^2-4|\geq \varepsilon[/itex]"

    So you don't need things to hold for all x such that [itex]|x-0|<\delta[/itex], but only for one specified x. Does that clear things up?
  4. Jul 13, 2011 #3
    Hello micromass, I appreciate the response!

    I'm getting confused here how to formalize this. So for all [itex]\delta[/itex], we can find such an x as you described. It makes sense in the terms of [itex]delta = 1[/itex], look at x= 1/2, with [itex] \epsilon = 1[/itex].
    I see that this case works, but how do do this for every delta? I can say now that I found an x for any [itex]\delta\geq 1[/itex] but now what about [itex]delta\leq 1[/itex] without going through the same argument?

    I guess we could let x= 1/n and say there exists an n by the archimedean property that this is true and therefore we can do it that way, but it seems to me I used to do this without using this step... but maybe I didn't?
  5. Jul 13, 2011 #4
    So you take [itex]\varepsilon=1[/itex]. For all [itex]\delta>0[/itex], you must find an x such that

    [tex]|x^2-4|\geq 1,~\text{but}~|x|<\delta[/tex]

    Indeed, if [itex]\delta >1[/itex], you can take x=1.
    And if [itex]\delta\leq 1[/itex], what if you take [itex]x=\delta/2[/itex]?
  6. Jul 13, 2011 #5
    That's a better idea. Then [itex]|\frac{\delta}{2}^{2}-4|\geq 1[/itex] since [itex]\delta/2\leq1/2[/itex]

    Would there be a way to do this without the caveat when [itex]\delta>1[/itex]. it seems very inelegant.
  7. Jul 13, 2011 #6
    Epsilon-delta stuff always tend to be inelegant :smile: But no, I don't think there is an easier/more elegant way of doing this.
  8. Jul 13, 2011 #7
    I think the way I'm thinking of it defining x=min(d/2,1). That would make me happy.

    Thank you so much for your help.
  9. Jul 13, 2011 #8
    Oh, yes, if you find that more elegant, then you can always do that of course :smile:
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