Proving where a multivariable function is continuous

**Pinkk** · August 31st 2010, 04:38 PM

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

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

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

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

Edit: Er, stupid me, meant $\frac{1}{2}x^{2}$ , not $2x^{2}$ , otherwise what I wrote doesn't make sense.

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