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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 $\lim_{x \to 0}{\sqrt{-|x|}}$

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    It is not possible to talk about the existence of$a$limit since there doesn't any sequence of points {$a_n$}$_{n\ge 1}$ in the domain of the function such that $a_n \ne 0$ and $a_n$ converges to zero, thus it is not possible to check if the image sequence has a limit or not.2015-11-23

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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 $\forall \epsilon>0\exists \delta>0:0<\left|x-a\right|<\delta\implies \left|f(x)-L\right|<\epsilon$

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    @Ord The "accumulation point" is added so that 0<\left|x-a\right|<\delta "makes sense", that is there exist $x$ so that it is true. In most definitions this is not explicitely written out, but I choose to do so to emphasise the importance of $a$ being$a$limit point2012-12-29
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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.