Results 1 to 4 of 4

Math Help - Quick question

  1. #1
    Senior Member slevvio's Avatar
    Joined
    Oct 2007
    Posts
    347

    Quick question

    Let f:\mathbb{R}^2 \rightarrow \mathbb{R} be a function such that along every line y=Ax and the line x=0, the function is continuous at (0,0). By this I mean if we let L_{\theta} be one of the lines in the x-y plane given by rotating an angle \theta anti-clockwise from the x-axis, and let \{ a_n \} be a sequence converging to (0,0) s.t. a_n\in L_{\theta} for all n, then f(a_n) \rightarrow f(0,0) as n \rightarrow \infty.

    Is it true then that f is continuous at (0,0)? So if we take an arbitrary sequence converging to (0,0), say \{ b_n \}, then does f(b_n) \rightarrow f(0,0)? I was thinking that the each element b_n lies on some line  L_{\theta_n} so that there are countably many lines that the sequence lies on, but I'm not sure if that means anything. Is the statement even true?

    Thanks for any help, just something that was bothering me
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Tinyboss's Avatar
    Joined
    Jul 2008
    Posts
    433

    Re: Quick question

    Hint: suppose the slope of f restricted to the line L_\theta is m_\theta, and suppose there's a sequence \theta_i such that m_{\theta_i}\to\infty.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721

    Re: Quick question

    If I understood this, take f(x,y)=\frac{xy^2}{x^2+y^4}. It has limit zero on every line through the origin, but if you approach zero through the parabola x=y^2 the limit is \frac{1}{2}
    Last edited by Jose27; July 8th 2011 at 05:44 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member girdav's Avatar
    Joined
    Jul 2009
    From
    Rouen, France
    Posts
    675
    Thanks
    32

    Re: Quick question

    Quote Originally Posted by Jose27 View Post
    but if you approach zero through the parabola x^2=y the limit is \frac{1}{2}
    Did you mean x=y^2 ?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. quick log question
    Posted in the Algebra Forum
    Replies: 7
    Last Post: January 9th 2009, 05:05 AM
  2. A quick question...
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: March 27th 2008, 12:22 PM
  3. quick question
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: January 6th 2008, 05:18 AM
  4. Just a quick question
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: March 8th 2007, 11:33 AM
  5. Quick's quick question
    Posted in the Number Theory Forum
    Replies: 22
    Last Post: July 9th 2006, 04:38 PM

Search Tags


/mathhelpforum @mathhelpforum