1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: 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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook