Is a function defined at a single point continuous?

Is a function defined at a single point continuous?

For example $f:\{0\}\to\{0\}$ defined by $f(x)=\sqrt{x}+\sqrt{-x}$ is a sum of two continuous functions and is therefore continuous, however for $f$ to be continuous, it has to have the limit $\lim\limits_{x_0\to x}f(x_0)=f(x)$ but it dosen't seem to have that limit.

Technically, yes. The domain is a singleton, so the only topology on it is the trivial one, $\tau_0$. Then if $(Y,\tau)$ is any other topological space and $*$ is the one-point space

$$f:(*,\tau_0)\to (Y,\tau)$$

is trivially continuous because if $U\in\tau$ is open

$$f^{-1}(U)\in\{\varnothing, *\}=\tau_0$$

depending on whether or not $f(*)\in U$ or not.

Just to address GitGud's objection, we note that this agrees with the naïve, calculus notion of a limit as well, we're required that all $x\in \{0\}\setminus \{0\}=\varnothing$ satisfy some property. Of course there are no counterexamples (the set is empty!) so it's true, because the definition of "truth" is "that for which no instance is false." There aren't any false instances in the null set (there aren't any at all!)

This in fact follows from the above, but for the student who may not be familiar with the modern definition of continuity (which certainly resolves this problem much more directly) this is another way to think of it.

And of course, even in calculus we routinely talk about continuity on non-open subsets of $\Bbb R$. The mean-value theorem, a hallmark of calculus, has as a hypothesis "continuous on $[a,b]$." If we are not willing to extend to singletons, this seems rather unfairly favoring of intervals.

Adam's answer is the most efficient way of looking at it, but here's another if you're interested.

You define continuity as $\lim\limits_{x_0\to x}f(x_0)=f(x)$ . Let's look at what this says: $$\forall\varepsilon > 0 \,\exists \delta \text{ s.t.} \forall x \in(x_0-\delta,x_0+\delta)\cap A\setminus\{x_0\}\, , \,\mid f(x_0) - f(x)\mid < \varepsilon$$

But, as you'll notice, $(x_0-\delta,x_0+\delta)\cap A\setminus\{x_0\} = \emptyset$, for any $\delta$. Now, the following is a bit of logic trickery, but it's like this: The hypothesis is false, so the implication statement is true. Every point in $(x_0-\delta,x_0+\delta)\cap A\setminus\{x_0\}$ satisfies that $\mid f(x_0) - f(x)\mid < \varepsilon $, because there are no such points to consider. It's something that took me quite some time to wrap my head around, but it's something one gets used to after proving many pathologically trivial statements.