**Well-known example:** Consider the function
$$f(x,y)=\left\{\begin{array}{c}
\frac{x^2y}{x^4+y^2} & \text{if }(x,y)\neq(0,0)\\
0 & \text{if }(x,y)=(0,0)
\end{array}\right.$$
When restricted to any straight line through the origin, this function is continuous. However, if we approach the origin along the parabola $y=x^2$, we get a limit of $\frac 12$, so $f$ is actually discontinuous. The question is whether smooth curves can always ferret out discontinuity in this way.

Does there exist a function $f:\mathbb R^2\to \mathbb R$ which is discontinuous at a point $x$, but is continuous at $x$ when restricted to any smooth curve?

Discontinuity is witnessed by a sequence $\{x_i\}$ converging to $x$ so that $\{f(x_i)\}$ does not converge to $f(x)$. Since we could take $f$ to be the characteristic function on $\{x_i\}$, the above question is equivalent to the following.

Given a convergent sequence $\{x_i\}$ in $\mathbb R^2$, must there be a smooth curve passing through infinitely many of the $x_i$?