TITLE: In higher dimensions (n>1) can jump discontinuities exist for otherwise continuous functions?
QUESTION [5 upvotes]: Real-valued functions mapping from $\mathbb{R}^1$ can exhibit jump discontinuities, even if they are continuous at every other point. For example, consider the function:
$$
f(x) =  
\begin{cases}
-1 & x < 0 \\
1 & x \geq 0
\end{cases}
$$
We say this function has a jump discontinuity at zero because approaching zero from the left yields a sequence with a different limit than approaching zero from the right:
$$\lim_{x \to 0-} f(x) \neq \lim_{x \to 0+} f(x).
$$
This jump discontinuity exists despite the fact that the function $f$ is continuous at every other point.
Now, let us consider a real-valued function $f$ mapping from $\mathbb{R}^n$ for $n > 1$ that is continuous at every point except possibly at $x = x_0$. I want to know if such a function can possibly have a jump discontinuity. That is, I'm wondering if there can exist a sequence of points $x^1_1, x^1_2,\dots$ and a sequence of points $x^2_1, x^2_2,\dots$ so that $\lim_{i \to \infty} x^1_i = x_0$ and $\lim_{i \to \infty} x^2_i = x_0$ but $\lim_{i \to \infty} f(x^1_i) \neq \lim_{i \to \infty} f(x^2_i)$.
For $n = 2$, we can use physical intuition by considering a bent sheet of paper, letting the height of the paper at each point represent the value of $f$. Even if we punch a (very small) hole in the piece of paper (representing some breakdown of continuity), we will still approach the same height as we approach the hole from any direction along the sheet of paper.
I am struggling to formalize this argument. I keep wanting to use the continuity of $f$, but the point of interest $x_0$ is precisely where this property (possibly) fails. Also, I know my argument needs to incorporate the dimension at some point (as jump discontinuities exist for dimension $n=1$), but it's not clear to me how to do this.
Any thoughts are appreciated.
Edit
In the comments, it was suggested to consider $f(x,y) = 2xy/(x^2+y^2)$, as explored at Continuity of $\frac{2xy}{x^2+y^2}$ at $(0,0)$. This is a ratio of continuous functions and so is continuous everywhere except at $(0,0)$ where it is undefined. Letting $y=ax$ for $a \neq 0$, we get $f(x,ax) = 2a/(1+a^2)$. We can approach the origin from different directions by setting $a$ to different values, and we will approach different values for $f$ at $(0,0)$ by doing this. Realizing this is possible is progress. However, this particular function is not defined at every point on $\mathbb{R}^2$. Is this sort of jump discontinuity still possible for functions defined at every point, so they really have $\mathbb{R}^n$ as their domain for $n > 1$?
Edit 2
It has been shown in the responses to this question that functions exhibiting this jump discontinuity exist at least in the case where $f$ is unbounded or limits of sequences of $f(x_i)$ as $x_i$ approaches $x_0$ are not guaranteed to exist. These examples exploit behavior impossible to achieve with a piece of paper, and so I'm still trying to pin down the physical intuition from that example. Can these jump discontinuities still exist if $f$ is required to be bounded, and the limits described above as one approach $x_0$ exist?
Edit 3
As pointed out in the comments, the function described in Continuity of $\frac{2xy}{x^2+y^2}$ at $(0,0)$ is indeed defined everywhere, as a piecewise definition is used to supply a value at the origin. (However, it is still true that $f(x,y) = 2xy/(x^2+y^2)$ is not defined everywhere). However, it remains unbounded [note: this is incorrect, as was pointed out in the comments - see Edit 4].
Edit 4
It was pointed out in the comments that $f(x,y) = 2xy/(x^2+y^2)$ (and say $f(0,0)=(0,0)$) is actually bounded. So, this provides an example of a function that exhibits the jump discontinuity behavior (meeting all the regularity criteria I wanted)!

REPLY [4 votes]: Here is  a whole collection of examples: let $g: [0,\infty) \to \mathbb R$ be any bounded function which is continuous on $(0,\infty)$ but does not have a limit as $x \to 0+$. Let $f(x)=g(\|x\|)$. Then $f$ has the required properties.