Can a limit of a function that's defined only in one point exist? The question is whether by definition there can exist a limit of a function that's defined only in one point ( or in several points but there's no interval in which the function is defined).
This came up when thinking about $\lim_{x \to 0}{\sqrt{-|x|}}$ 
 A: No. For the limit $\lim_{x\to a}f(x)$ to have a meaning $x$ will have to be able to go as close to $a$ as desired. Formally if $f:X\subseteq \mathbb{R}\to \mathbb{R}$ we demand that $a$ is an accumulation point of $X$, that is:
$$\forall \epsilon>0\ \exists x\in X:\ 0<\left|x-a\right|<\epsilon$$
which in our case is not true. The point $0$ is isolated in $\left\{0\right\}$ and so the symbol $\lim_{x\to a}f(x)$ has no meaning. 
You may want to note that $f$ is however continuous at $0$, in fact any function is continuous at the isolated points of its domain.
EDIT: To be unambiguous here is my definition of limit:
If $f:X\subseteq \mathbb{R}\to \mathbb{R}$ and $a$ is a limit point of $X$ then $\lim_{x\to a}f(x)=L\in \mathbb{R}$ if
$$\forall \epsilon>0\exists \delta>0:0<\left|x-a\right|<\delta\implies \left|f(x)-L\right|<\epsilon$$
A: A limit of a real-valued function defined only at one point does not exist, such as wikipedia defines limits. The limit value of a function at a point $c$ is defined from the function values of points arbitrarily close to, but not equal to $c$.
An example would be the function that has $f(0) = 1$ and is $0$ everywhere else. Then the limit value should be defined so that it is $0$ at every point. Thus the definition of limit at $0$ cannot take the function value at $0$ into account.
