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Math Help - f is not continuous

  1. #1
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    f is not continuous

    Hello everyone

     f: \mathbb{R}^2 \to \mathbb{R}, \ \ \ x,y \in \mathbb{R}

    f(x,y) := \begin{cases} \frac{xy}{x^2+y^2}, & \mbox{if } (x,y) \not= (0,0) \\ 0, & \mbox{if } (x,y) = (0,0) \end{cases}

    Show that f is not continuous in (0, 0)

    I really do not know how to solve that.

    Thanks for your time.

    Rapha
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by Rapha View Post
    Hello everyone

     f: \mathbb{R}^2 \to \mathbb{R}, \ \ \ x,y \in \mathbb{R}

    f(x,y) := \begin{cases} \frac{xy}{x^2+y^2}, & \mbox{if } (x,y) \not= (0,0) \\ 0, & \mbox{if } (x,y) = (0,0) \end{cases}

    Show that f is not continuous in (0, 0)

    I really do not know how to solve that.

    Thanks for your time.

    Rapha
    In order to show it is continuous, you must show that \lim_{(x,y) \to (0,0)} \frac{xy}{x^2+y^2}=0, for any direction.
    Since you want to show it is not continuous, it is sufficient to find a direction for which the limit is not 0.

    Switch to polar coordinates. The limit will become :
    \lim_{r \to 0} \frac{r^2 \cos \theta \sin \theta}{r^2}=\lim_{r \to 0} \cos \theta \sin \theta

    which is not always equal to 0.

    Hence it is not continuous at (0,0)
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  3. #3
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    Try two different paths to (0,0). Say y=x and y=-x.
    Do you get different limits?
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  4. #4
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    Quote Originally Posted by Moo View Post
    Hello,

    ...
    Merci beaucoup.

    Quote Originally Posted by Plato View Post
    Try two different paths to (0,0). Say y=x and y=-x.
    Do you get different limits?
    Thank you very much. I get different limits. Good stuff :-)
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