Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Check work on 2 variable function.

  1. Sep 26, 2009 #1
    1. The problem statement, all variables and given/known data

    Determine the set of points at which the function is continuous.

    F(x,y) = R.

    where R is a piecewise function of :

    x^2*y^3 / (2x^2 + y^2) ; if(x,y) != (0,0)
    1 ; if(x,y) = (0,0)

    Obviously, the first function is not defined at point (0,0), but to find
    the domain of the piecewise function, I first need to see if the
    first function is at least continuous.

    So here is my attempt at that :

    Let A = x^2 * y^3 / (2x^2 + y^2 )

    the |A| is =

    x^2 * |y^3|
    2x^2 + y^2

    well x^2 <= 2x^2 + y^2, lets call that J

    so A < J * |y^3| / (J) = |y^3| = sqrt(y^6), and we see that this function
    is defined at point 0 , thus lim of A as (x,y) -->(0,0) = 0. ???

    So if the above is true then the peicewise function should be
    defined in region R^2???

    I am not sure if this is correct. The book says that the answer is :

    { (x,y) | (x,y) != (0,0) }.

    I think that means the function A is not defined at 0 thus the peicewise
    function is not defined at point (0,0). What did I do wrong ?
  2. jcsd
  3. Sep 27, 2009 #2
    You showed [tex]\lim_{(x,y)\to (0,0)} F(x,y)=0[/tex] . However, since F(0,0)=1 by the definition of F, we have that F is not continuous at (0,0).
  4. Sep 27, 2009 #3
    I thought that given value defined in the piecewise, 1 in this case, was an arbitrary value.
    So it did not matter, unless the limit of part A in the piecewise function did not exist?
  5. Sep 27, 2009 #4
    If F(0,0) had been defined to be 0 instead of 1, then F would have been continuous everywhere.

    If F(0,0) had been defined to be c, with c nonzero (c=1 is a special case), then as in your problem, F would not have been continuous at (0,0) but would be continuous everywhere else.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Check work on 2 variable function.