What is the point $(0,0)$ of $\sqrt{x}$ called? $\sqrt{x}$ is defined over $\left [ 0,+\infty  \right ]$
What's confusing me is that my teacher used to call the point $(0,0)$ of this function , a point of discontinuity, but this function is continuous over its domain
 A: I note the various answers here all discuss the continuity (which it is), but do not answer the question in the title - which is, what is the best name by which to call this point? To that end, I would say that the best name for it is probably the endpoint of the graph of $\sqrt{}$. And "graph" here is actually a strictly-defined mathematical object, not simply the thing you draw on paper, which is really a picture of the graph (cf. Rene Magritte's famous painting, The Treachery of Images). The graph of a function $f$ is the set of all ordered pairs $(x, y)$ such that $y = f(x)$, and can be considered as one of the three ingredients that define a function, the other two being the domain and codomain (though you could argue that only two are really required and one is redundant: the domain can be recovered from the graph, as the set of all the first elements of the ordered pairs).
In this case, the graph inherits a topology from the plane, and is topologically equivalent to the half-open interval $[0, \infty)$ (think of the curve as just a "bent half-line"). The point $0$ on that interval corresponds to $(0, 0)$ on the graph of $\sqrt{}$, hence since we call $0$ an endpoint of such interval, I'd call such an endpoint of the graph as well.
A: Let $f(x)=\sqrt{x}$ for $x \ge 0.$ Then:
$$ \lim_{x \to o}f(x)=0=f(0).$$
Consequence: $f$ is continuous at $0$.
A: Topological answer.
If you are not familiar (yet) with topology then this answer might be quite useless for you. Nevertheless I publish it to promote completeness.

If $f$ is a function $X\to Y$ where domain and codomain are both underlying sets of topological spaces then $f$ is (according to the common definition in topology) continuous at $x_0\in X$ iff:


*

*For every open set $U\subseteq Y$ with $f\left(x_0\right)\in U$ there
is an open set $V\subseteq X$ such that $x_0\in V$ and $f\left(V\right)\subseteq U$
Applying that here on $X=[0,\infty)$ and $Y=\mathbb R$ and $f$ prescribed by $x\mapsto\sqrt{x}$ where the sets are equipped with their usual topologies we find that $f$ is continuous at $0$.
A: $f$ is indeed continuous at $x=0$, as Fred already explained.
$f$ is not differentiable at 0, though, as $$\lim_{h \to 0^+} \frac{\sqrt{0+h}- \sqrt{0}}{h}$$ does not exist! Also,  $f(x):=\sqrt{x}$, then $$\lim_{x \to 0} f'(x) = +\infty$$
If you are not studying pure mathematics, but rather physics or any other science, you teacher may have taken that license at call $x=0$ a "singularity", as $f$ is not "that smooth" there as it is in the rest of its domain
