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.