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Homework Help: Multivariate ε-δ proof

  1. Aug 10, 2012 #1


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

    Show that : (x^4+y^4)/(x^2+y^2) < ε if 0 < x^2 + y^2 < δ^2 for a suitably chosen δ depending on ε.

    2. Relevant equations

    [itex]\forall[/itex]ε>0, [itex]\exists[/itex]δ>0 | 0 < (x^2 + y^2)^(1/2) < δ [itex]\Rightarrow[/itex] |f(x,y) - L| < ε

    Obviously here were dealing with lim (x,y)→(0,0) f(x,y) = 0 so the following statement is equivalent and more convenient to use in my opinion:

    [itex]\forall[/itex]ε>0, [itex]\exists[/itex]δ>0 | 0 < |x|,|y| < δ [itex]\Rightarrow[/itex] |f(x,y) - L| < ε

    3. The attempt at a solution

    So we know : |x| < δ [itex]\Rightarrow[/itex] x^2<δ^2 and also |y|<δ [itex]\Rightarrow[/itex] y^2<δ^2

    And using the triangle inequality we also consider : |x^4 + y^4| ≤ |x|^4 + |y|^4

    So putting those together we observe :

    |f(x,y) - L| = |(x^4+y^4)/(x^2+y^2)| ≤ (|x|^4 + |y|^4)/(|x|^2 + |y|^2) < 2δ^4/2δ^2 = δ^2 ≤ ε

    [itex]\Rightarrow[/itex] δ = [itex]\sqrt{ε}[/itex]

    Now that I have my δ, I could go through and prove that it was the right δ, but I have one problem. The book says that δ = [itex]\sqrt{ε/2}[/itex] so I'm wondering where I went wrong or is this a typo in the book? If it helps I also tried using the other statement 0 < (x^2+y^2)^(1/2) < δ and got the right answer, but I'm not sure why I'm wrong about this other method?

  2. jcsd
  3. Aug 10, 2012 #2
    Although I agree with your choice of [itex]\delta=\sqrt{\epsilon}[/itex], I disagree with some of your calculations.

    Your statement of limit is incorrect here. If should read [itex]0< ||(x,y)||<\delta \Rightarrow |f(x,y)|<\epsilon.[/itex]

    Where I have taken your L to be zero, as is the question.

    This is true, but it is not necessary that we require |x| < δ and |y| < δ since this follows whenever [itex]||(x,y)||<\delta[/itex]. Again, this is just a clarification. I more so wish to emphasize that whatever limit definition you used above was incorrectly copied down.

    This is fine. However, triangle inequality is not required since x^4 and y^4 are both positive.

    The second to last inequality does not follow. In general, we have the following:
    [tex] (x^2+y^2)<\delta \Rightarrow 1/\delta<1/(x^2+y^2). [/tex]
    However, you have (falsely) assumed
    [tex] (x^2+y^2)<\delta \Rightarrow 1/(x^2+y^2)<1/\delta[/tex]
    or something of that form.

    A quick way to prove the end result would be to, first, drop those abolute values (since |x|^2=x^2, etc.), and then notice that [itex]x^4+y^4=(x^2+y^2)^2-2x^2y^2[/itex], and work from there.
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