does there exist a function of two variables that is not continuous at origin, but whose restriction to every polynomial is, Does there exist a function from $R^2$ to $R$ that is not continuous at origin, but whose restriction to every single polynomial curve through the origin is continuous. 
I am not too sure how to approach this problem. I tried just creating some functions that I thought would have said property and then trying to find paths for them (such as $sin(x)$ or $e^x - 1$) that would contradict it. I know this isn't a good approach because there is no way to test every single polynomial, so there must be something I am missing. My initial guess was that it is impossible since any function can at least be approximated to some degree using a polynomial, but once again I know that isn't a valid proof
Any help would be appreciated, I am really stumped
Thanks
 A: Yes, there exists such a function.
Consider, for example,
$$
f(x,y) :=
\begin{cases}
1, & \text{if}\ y = |x|^{3/2}, \ x\neq 0,\\
0, & \text{otherwise}. 
\end{cases}
$$
A. Rosenthal proved that (see here):
If a function $f\colon \mathbb{R}^2\to\mathbb{R}$ is continuous at a point $P$
along every convex curve through $P$ which is (at least) once differentiable, then it is continuous at $P$.
Yet, $f$ can be continuous at $P$ along every curve which is (at least) twice differentiable without being continuous at $P$.
A: You have not defined a "polynomial curve". Im considering curves of the form
$$\gamma:\quad t\mapsto{\bf z}(t)=\bigl(p(t),q(t)\bigr)\qquad(-\infty<t<\infty)\tag{1}$$
with $p$ and $q$ polynomials,  $\>{\bf z}(0)={\bf 0}$ and  $\>{\bf z}'(0)\ne{\bf 0}$. (This  includes graphs $y=q(x)$, $q$ a polynomial.)
The function I propose is
$$f(x,y):=\left\{\eqalign{1\quad&\bigl(x\ne 0\ \wedge\ 0<|y|<e^{-1/x^2}\bigr)\cr 0\quad&({\rm otherwise})\ .\cr}\right.$$
The reasons this works is the following: Any curve $\gamma$ considered in $(1)$ has a well defined tangent at ${\bf 0}$. If this tangent is not horizontal then $f\bigl({\bf z}(t)\bigr)\equiv0$ for small $|t|$. If this tangent is horizontal then either $y(t)\equiv0$, or there is an $r\geq2$ with $|y(t)|\geq Cx^r$ for small $|t|$. One then can check that $f\bigl({\bf z}(t)\bigr)\equiv0$ for small $|t|$ as well.
