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Continuous extension(2)

  1. Aug 11, 2010 #1
    How can the function

    f: ℝ² → ℝ : (x,y) |--> [tex]{{x^3-y^3}\over{x-y}}[/tex] if x ≠ y

    be defined on the line y=x so that we get a continuous function?


    Is this correct?: If x=y --> f=0
     
  2. jcsd
  3. Aug 11, 2010 #2

    HallsofIvy

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    Science Advisor

    No. Did you have any reason at all for thinking that?

    For [itex]x- y\ne 0[/itex]
    [tex]\frac{x^3- y^3}{x- y}= x^2+ xy+ y^2[/itex]

    As long as x is NOT equal to y, that function is the same as [itex]x^2+ xy+ y^2[/itex]. The function will be continuous on the line the line x= y, if we define [itex]f(x, y)= f(x, x)= x^2+ xy+ y^2= x^2+ x^2+ x^2= 3x^2[/itex], not "0".
     
  4. Aug 11, 2010 #3
    Thank you very much!
     
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