1. Not finding help here? Sign up for a free 30min 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!

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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Prove a function is not continuous
Loading...