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Function continuous or not at (0,0)

  1. Jan 24, 2012 #1
    1. The problem statement, all variables and given/known data

    [itex]f\rightarrowℝ[/itex], [itex]f(x,y)=\frac{x^{2}y}{x^{6}+y^{2}}[/itex] where [itex](x,y)\neq(0,0)[/itex] and [itex]f(0,0)=0[/itex].

    Is the function continuous at [itex](0,0)[/itex]?

    3. The attempt at a solution

    I tried to find the limit at [itex](0,0)[/itex] so I put [itex]y=x[/itex] into the function [itex]f[/itex] and got the limit 0 when [itex]x\rightarrow0[/itex]. Tthen I put [itex]y=x^{2}[/itex] into [itex]f[/itex] and got the limit 1 when [itex]x\rightarrow0[/itex]. That means that the limit does not exist right?
    But the part that says [itex]f(0,0)=0[/itex] confuses me. Does that change the limit?

    There is a second part for this problem where I'm supposed to find the first partial derivatives in [itex](0,0)[/itex] or explain why they do not exist but I'd like to understand this first and then try to see if I can do the second part by myself. I think that if the limit does not exist in [itex](0,0)[/itex] then the partial derivatives can not either by definition... But I'm not sure...

    Thank you in advance. :smile:
     
  2. jcsd
  3. Jan 24, 2012 #2

    SammyS

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    You asked:
    "But the part that says [itex]f(0,0)=0[/itex] confuses me. Does that change the limit?"​
    The answer is no!

    The limit as (x,y) → (0.0) has nothing to do with the value of f(0,0). Indeed, this limit can exist even if f(0,0) is undefined.

    By The Way: Your solution to this problem is correct. You have shown that the limit does not exist.
     
  4. Jan 26, 2012 #3
    Thank you. :smile:
     
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