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Consider a function defined on $\mathbb{R}^2$ by

$$f(x,y) = \begin{cases} 0 & \quad \text{if $y\leq{0}$ or if $y\geq{x^2}$}\\ \sin\left(\frac{\pi y}{x^2}\right) & \quad \text{if $0<y<x^2$} \end{cases}$$

(a) Show that $f$ is not continuous at the origin. (WITHOUT using limit)

(b)Show that the restriction of $f$ to any straight line through the origin is continuous


So, I drew the graph.

For $f(x,y) = 0$, It is everything above the $y=x^2$ parabola and everything below the line $y=0$

For $f(x,y) = \sin(\frac{\pi y}{x^2})$, its the area of everywhere else

(a) I dont get it, the graph $f(x,y)$ actually contains the origin on $f(x,y) = 0$ which is part the parabola and the line

(b) I used the fact that the graph of $f(x,y) = 0$ actually cover everything on or under $y=0$, so no matter what straight line with whatever slope I draw, part of the line will always fall onto that region. Thats why it is continuous.

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  • $\begingroup$ What do you mean by "without using limit"? $\endgroup$
    – Pedro
    Mar 10, 2013 at 23:33
  • $\begingroup$ Cannot use the method on looking into how the curve behave near the origin $\endgroup$
    – Paul
    Mar 10, 2013 at 23:34

1 Answer 1

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For (a), the origin $(0,0)$ doesn't satisfy $0<y<x^2$ so only the top of the piecewise definition applies, so that $f(0,0)=0.$

Your argument for (b) is basically correct.

To finish showing discontinuity, note that (except at $(0,0)$) the curve $y=x^2/2$ satisfies $0<y<x^2$, so that the second rule applies. So at these points $$f(x,y)=\sin([\pi \cdot x^2/2]/[x^2])=\sin(\pi/2)=1,$$ and this is without taking a limit, and gives a jump as one passes through $(0,0)$ along $y=x^2/2$ of a sudden $1$ unit downwards as $(0,0)$ is crossed, since $f(0,0)=0.$

Just noticed your comment that you "can't use looking at the curve near the origin". But one has to look near the origin, if one is to show discontinuity there...

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