TITLE: Is a function defined at a single point continuous?
QUESTION [13 upvotes]: 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.

REPLY [2 votes]: 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.