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Defining the continuity and differentiability of multi variate functio

  1. Jun 6, 2013 #1
    Let f: R2-->R be defined by f(x,y) = xy2/(x2+y2 if (x,y) ≠ 0, f(0,0) = 0

    a) is f continuous on R2?
    b) is f differentiable on R2?
    c) Show that all the dirctional derivatives of f at (0.0 exist and compute them

    Attempt:

    a) I had an idea to show that multivariate functions are continuous, but now I think it's faulty. I did this:

    xy2/(x2+y2 ≤ |xy2|/|(x2+y2| < |xy2|

    now taking the lim (x,y)-->(0,0) of |xy2| is 0 as such the function is also continuous.

    But looking at a similar example from my notes where the only difference in the function was:


    xy2/(x2+y6

    We had shown that this was not continuous. So now I'm not so sure if the method I used was valid.

    b) I'm a little stumped on showing differentiability in this setting, it doesn't seem to involve using the definition directly because of all the possible paths. So what other option may I have?

    This is all in terms of discussing the trouble spot. i.e approaching (0,0)
     
  2. jcsd
  3. Jun 6, 2013 #2
    Find the limit when ##x_n = y_n = \frac 1 n ##, and when ##x_n = y^2_n = \frac 1 n ##.
     
  4. Jun 6, 2013 #3
    I dont follow. By doing that i get different solutions taking n-->0.
     
  5. Jun 6, 2013 #4

    Zondrina

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    Homework Helper

    So you have :

    ##f(x,y) = \frac{xy^2}{x^2 + y^2}## if ##(x,y) ≠ (0,0)##
    ##f(x,y) = 0## if ##(x,y) = (0,0)##

    For part a) where you're asked if ##f## is continuous, this should ring bells about the ##ε - δ## definition if you want to show it rigorously. Notice you can already tell that the function is continuous everywhere ( since it is a quotient of polynomials and polynomials are continuous ) except at (0,0). So using the definition with the information we're given :

    ##\forall ε > 0, \exists δ > 0 \space | \space 0 < |(x,y) - (0,0)| < δ \Rightarrow |\frac{xy^2}{x^2 + y^2} - 0| < ε##

    So now you want to massage the expression into a suitable form :

    ##|f(x,y) - L| = \frac{|x|y^2}{x^2 + y^2}##

    Now, using the fact that ##|x| < δ## and ##|y| < δ## you should be able to continue.
     
  6. Jun 6, 2013 #5
    n was supposed to be integer, going to infinity. Does that make (x, y) go to zero?
     
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