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: 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


    User Avatar
    Science Advisor
    Homework Helper

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook