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Help with epsilon-delta definition of a limit of a function

  1. Jun 17, 2009 #1
    1. The problem statement, all variables and given/known data


    [tex]f(x,y) = \dfrac{x^2+2xy^2+y^2}{x^2+y^2}[/tex]

    Prove that

    [tex]\lim_{(x,y) \to (0,0)} f(x,y) = 1[/tex]

    2. Relevant equations

    Definition of the limit of a function of multiple variables:

    It suffices to show that for all [tex]\epsilon > 0[/tex], there exists a [tex]\delta > 0[/tex] such that for all [tex](x,y)[/tex] such that [tex]0 < x^2 + y^2 < \delta ^2[/tex], we have [tex]|f(x,y) - 1| < \epsilon[/tex]

    3. The attempt at a solution

    [tex]|f(x,y) - 1| = \left| \dfrac{2xy^2}{x^2+y^2} \right| = \dfrac{2|x|y^2}{x^2+y^2} [/tex]

    I need to bound this with an expression in terms of [tex]\delta[/tex], but I can't think of any way to do so. I noticed that the denominator is less than [tex]\delta ^2[/tex] but I can't get anywhere with that. (I end up bounding it in the wrong direction! :uhh:)

    Can anyone point me in the right direction? Thanks.

    [Edit: Good catch Mark44!]
    Last edited: Jun 18, 2009
  2. jcsd
  3. Jun 17, 2009 #2


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    Try 2|x|y^2<=2|x|*(x^2+y^2). How can you bound |x| in terms of delta?
  4. Jun 17, 2009 #3


    Staff: Mentor

    This isn't much help, but the denominator is less than [itex]\delta ^2[/itex], which makes the overall expression larger than it would be if the denominator were equal to [itex]\delta ^2[/itex].
  5. Jun 18, 2009 #4
    I get that [tex]|x| = \sqrt{x^2} \leq \sqrt{x^2 + y^2} < \delta[/tex] so

    [tex]\dfrac{2|x|y^2}{x^2+y^2} \leq \dfrac{2|x|(x^2 + y^2)}{x^2+y^2} = 2|x| < 2\delta[/tex]

    So for any [tex]\epsilon[/tex], we can choose [tex]\delta = \epsilon/2[/tex].

    Thanks Dick! That was tricky.
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