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Prove a function is not continuous

  1. Nov 16, 2009 #1
    1. The problem statement, all variables and given/known data
    Using the epsilon-delta definition, prove that the function f:R[tex]^{2}[/tex] [tex]\rightarrow[/tex] R by f(x,y) = xy/((x[tex]^{2}[/tex]) + (y[tex]^{2}[/tex])), and f(0,0) = 0 is not continuous.

    3. The attempt at a solution
    I just really have no clue how to set up a delta-epsilon proof for functions that involve quotients. I went ahead and set up as much delta information as I could, but I have no idea how to set up the epsilon part:

    |x-x[tex]_{0}[/tex]|<[tex]\delta[/tex], |y-y[tex]_{0}[/tex]|<[tex]\delta[/tex], and |(x-x[tex]_{0}[/tex])+(y-y[tex]_{0}[/tex])|<[tex]\delta[/tex] (Those are supposed to be x (initial) and y (initial) for the delta info.... I couldn't get LaTex to set them up correctly...)

    Can someone give me a couple of good pushes in the right direction? :-)
  2. jcsd
  3. Nov 16, 2009 #2


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    For problems like these, I recommend changing to polar coordinates. That way, (x,y)-> (0,0) becomes just r-> 0 no matter what [itex]\theta[/itex] is. To show that this function is not continuous you only have to show that [itex]|f(r, \theta)|[/itex] cannot be made arbitrarily small ("[itex]< \epsilon[/itex]") for small r for at least some values of [itex]\theta[/itex]. For this particuar problem I think you will find that easy.
  4. Nov 16, 2009 #3
    Hm, I'm just not sure why I didn't think of that before! Thank you!
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