Proving continuity of a $\mathbb{R^2}\rightarrow \mathbb{R}$ function at a point.

Question

Let $f(x,y): \mathbb{R^2}\rightarrow \mathbb{R}$ with $f(x_0,y_0)=a$. To prove that $f$ is continuous at $(x_0,y_0)$ when approaching the point in the domain along a straight path, we define $g(t)=\begin{bmatrix}x_0\\y_0\end{bmatrix} + t\cdot \begin{bmatrix}r_x\\r_y\end{bmatrix}$, $(r_x,r_y)\neq(0,0)$ and $h(t) = f(g(t))$. Then if $\lim_{t\rightarrow 0}{h(t)}=a$, the statement is proven.

Question: is it possible for $f$ to still not be continuous at $(x_0,y_0)$? For example, if approaching the point along $t\cdot \sin{\frac{1}{t}}$ or other path that has no defined direction at $(x_0,y_0)$, or for other reasons.

Answer 1

Yes, $f$ may not be continuous. For example, consider the following $f$ given in polar coordinates where $0\leq\theta<2\pi$: $f(r,\theta) = \frac{1}{2\pi-\theta}r$. This is continuous on every straight path through the origin, but is not continuous in $\mathbb{R}^2$ at the origin because $\frac{1}{2\pi-\theta}r$ gets arbitrarily large as $\theta \to 2\pi$ for any $r$.

Answer 2

The typical example of this is $$ f(x, y) = \cases{\dfrac{x^2y}{x^4+y^2} & if $(x, y)\neq (0,0)$\\0 & otherwise} $$ If we go to the origin along the $x$ or $y$-axis, then the limit is clearly $0$. If we go along the line $y = ax$ with $a\neq 0$, then again we get that the limit is $0$, since $f(x, ax) = \frac{ax^3}{x^4 + a^2x^2} = \frac{ax}{x^2+a^2}$. But if we instead of a line go along the parabola $y = x^2$, we get $f(x, x^2) = \frac{x^4}{x^4 + x^4} = \frac12$.