Is it possible for a function to be continuous at a point if the function is not defined either to the left or right of that point? Take for example the function $f:[0,\infty) \to \Bbb{R}$ given by $f(x)=\sqrt x$ is this function continuous at $x=0$?
For it to be continuous we look for $$\lim _{x\to0} f(x)$$ and we say it is continuous if that limit is $f(0)=0$ but is $$\lim _{x\to 0} \sqrt x=0$$ I believe it is but can't we say if this limit exists then the left hand limit and right hand limit both exist and are equal but here the left hand limit does not exist since it is not in the domain of $f$ when approaching from the left?
Could anyone clear up the confusion?
 A: This is a very good question. The answer is that it depends on exactly how continuity is defined. Some people define it for real functions such that it must be defined on an open interval around the point, otherwise it is considered non-continuous. But a more general definition is:
$\def\wi{\subseteq}$

Given any metric spaces $S,T$, a function $f : S \to T$ is continuous at $x \in S$ iff, for any open set $B \wi T$ such that $f(x) \in B$, for some open set $A \wi S$ with $x \in A$ we have $Im_f(A) \wi B$.

Note that under this definition we would consider the square-root function on the non-negative reals to be continuous at $0$. Issues of left-continuity and right-continuity don't matter because they are irrelevant in the metric space sense.
A: In general topology $f:[0,\infty) \to \Bbb{R}$ defined by $f(x)=\sqrt x$ is a continuous function, the point outside the domain $[0,\infty)$ does not affect the continuity of $f$ in any way.
However, in pre-calculus the situation is more complicated. One often sees the definition of continuity as a function satisfying the following conditions:


*

*$f(x_0)$ exists.


*$\lim_{x\to x_0} f(x)$ exists


*The two are equal

The problem often lies with 2, since students are taught that $\lim_{x\to x_0} f(x)$ exists iff $\lim_{x\to x_0^+} f(x)$ and $\lim_{x\to x_0^-} f(x)$ exists - and in this case $\lim_{x\to x_0^-} f(x)$ doesn't.
I'd say that it depends on the convention you are using and how you are taught in the class. If it's a metric space or topology class then yes, it is continuous. However, if you are in an elementary calculus class and your teachers are somewhat unreasonable, then you might very well lose some points for saying that $f$ is continuous.
