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Thread: Proving where a multivariable function is continuous

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    Senior Member Pinkk's Avatar
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    Proving where a multivariable function is continuous

    Let $\displaystyle f(x,y) = y(y-x^{2})/x^{4}$ if $\displaystyle 0 < y < x^{2}$ and $\displaystyle f(x,y) = 0$ otherwise. Prove that $\displaystyle f$ is discontinuous only at $\displaystyle (0,0)$.

    So I am fairly certain I know how to show the function is discontinuous at $\displaystyle (0,0)$:

    If $\displaystyle f$ was continuous at $\displaystyle (0,0)$, then $\displaystyle \lim_{(x,y)\to (0,0)} f(x,y) = f(0,0) = 0$. However, if we approach $\displaystyle (0,0)$ along the parabola $\displaystyle y= 2x^{2}$, then we have that $\displaystyle \lim_{(x, 2x^{2})\to (0,0)} 2x^{2}(2x^{2} - x^{2})/x^{4} = \lim_{(x, 2x^{2})\to (0,0)} 2 = 2$ since $\displaystyle 0 < 2x^{2} < x^{2}$ as $\displaystyle (x, 2x^{2})$ gets sufficiently closer and closer to the origin. Therefore, $\displaystyle f$ is discontinuous at $\displaystyle (0,0)$.

    Is that proof correct, and if not, what is wrong?

    Moving to the next portion, I am stuck; how do I prove that for all other points, $\displaystyle f$ is continuous. I have a general idea that I need to check points outside the region $\displaystyle 0 < y < x^{2}$, inside that region, and then on the boundaries of the region (namely points along $\displaystyle y= x^{2}$ and $\displaystyle y=0$. I am having trouble formalizing this into a proof though. Thanks for any help.

    Edit: Er, stupid me, meant $\displaystyle \frac{1}{2}x^{2}$, not $\displaystyle 2x^{2}$, otherwise what I wrote doesn't make sense.
    Last edited by Pinkk; Aug 31st 2010 at 05:20 PM.
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