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Continuity of a Function of Two Variables

  1. Nov 8, 2004 #1
    Question:
    Is the function f(x,y) = (x^2 - y^2)/(x-y) continuous at (1,1) if we set f(1,1) = 0? Why or why not?

    So far, I've just plugged 1 in for x and y and found the limit to equal 0. I guess that means that the limit is not continuous at (1,1)? And what do they mean by set f(1,1) = 0?

    Thanks for your help!
     
  2. jcsd
  3. Nov 8, 2004 #2

    Galileo

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    The function f(x,y) = (x^2 - y^2)/(x-y) is not defined at (1,1), since:
    (1-1)/(1-1)=0/0.
    So setting f(1,1)=0, means defining a value for the function at (1,1).

    By the way. The limit is not equal to 0.
    Note that : (a+b)(a-b)=a^2-b^2.
     
  4. Nov 8, 2004 #3
    [tex]\lim_{(x,y)\rightarrow(1,1)}f(x,y)=\lim_{(x,y)\rightarrow(1,1)}\frac{x^2-y^2}{x-y}=\lim_{(x,y)\rightarrow(1,1)}\frac{(x+y)(x-y)}{x-y}=\lim_{(x,y)\rightarrow(1,1)}x+y=2[/tex]
     
    Last edited: Nov 8, 2004
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