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Epsilon/Delta Proof With 2 Variables

  1. Dec 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove:
    f(x,y) = [itex]\frac{x(x^{2}-y^{2}}{(x^{2}+y^{2}}[/itex] if (x,y) [itex]\neq[/itex] (0,0)
    0 if (x,y) = (0,0)

    is continuous at the origin
    2. Relevant equations

    [itex]\forall[/itex] [itex]\epsilon[/itex] > 0 [itex]\exists[/itex] [itex]\delta[/itex] > 0 s.t. if |(x,y)| < [itex]\delta[/itex] then |f(x,y)| < [itex]\epsilon[/itex]

    (Since we are proving continuity at the origin)

    3. The attempt at a solution

    |(x,y)| < [itex]\delta[/itex] [itex]\Leftrightarrow[/itex] x[itex]^{2}[/itex] + y[itex]^{2} < \delta^{2}[/itex]

    then this means that |x[itex]^{2}[/itex] - y[itex]^{2}[/itex]| < [itex]\delta^{2}[/itex]

    so:

    f(x,y) < [itex]\frac{x}{x^{2}+y^{2}}[/itex]([itex]\delta^{2}[/itex])

    and I feel like I'm close but then I'm stuck! All help appreciated thanks !
     
  2. jcsd
  3. Dec 10, 2012 #2

    Dick

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    Have you tried thinking about it in polar coordinates? It's much easier.
     
  4. Dec 10, 2012 #3
    hmm I have never used polar coordinates with an epsilon delta proof before

    so x=rcosθ
    and y=rsin

    so f(x,y) is rcosθ(cos[itex]^{2}[/itex]θ - sin[itex]^{2}[/itex]θ)

    and r<[itex]\delta[/itex] and cosθ <= 1

    so f(x,y) < [itex]\delta[/itex](cos[itex]^{2}[/itex]θ - sin[itex]^{2}[/itex]θ)

    which = [itex]\delta[/itex]cos(2θ) <= [itex]\delta[/itex](1) = [itex]\delta[/itex]

    set [itex]\delta[/itex] = [itex]\epsilon[/itex]

    does this work?
     
  5. Dec 10, 2012 #4

    Dick

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    Sure does. If r<ε then |f(r,θ)-f(0,0)|<ε.
     
  6. Dec 10, 2012 #5

    SammyS

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    Isn't that |f(r,θ)-f(0,θ)| < ε ?
     
  7. Dec 10, 2012 #6

    Dick

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    Oh, you know what I mean. But I don't see why you'd have to write it that way. f(0,θ)=0 and f(0,0)=0. Same thing. They are both the origin in xy coordinates.
     
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