can we say that $\lim\limits_{x\to0} \sqrt{-x^2}=0$ even if the function domain is a singleton $\{0\}$ In Real analysis, the strict definition of a limit, we take a sequence of point that converging to the point a "limit point". which is not verified here in this case we have only one element in the function domain :
$$\lim_{x\to 0} \sqrt{-x^2}$$
but by replacing $0$ in the function we get $0$
so my question is : in a formal way can we say that
$$\lim_{x\to 0} \sqrt{-x^2}=0$$
 A: We can say so with perfectly good conscience. Ignoring all the notational baggage, the function in question is
$$f:\{0\}\longrightarrow\mathbb R,~0\mapsto0.$$
Here, $\{0\}$ is a metric space with the only possible metric on a singleton set: $d(0,0)=0$. With this metric, the set $\{0\}$ gains the discrete topology, with respect to which every function to any other metric space (or even topological space) is continuous. It's not a particularly interesting topology, but it's a perfectly fine one.
Another argument why it's perfectly fine to allow such a function to be continuous: Let $f:X\longrightarrow Y$ be a constant map where $X,Y$ are metric spaces (or even topological ones). Then $f$ is continuous. This is a very intuitive statement which is also easy to prove in the context of topology. Your function is a constant function between metric spaces, so we'd have to make an exception for this theorem, which in my opinion would be way worse than calling your function continuous.
A: From the definition of a limit in a metric space:
$$
\forall\epsilon\gt0,\exists\delta\gt0:0\lt|x-0|\le\delta\implies|f(x)-0|\le\epsilon\tag1
$$
Since there are no $x$ in the domain of $f(x)=\sqrt{-x^2}$ that satisfy $0\lt|x-0|\le\delta$, the implication is vacuously true.

However...
Thanks to Adam Rubinson for pointing out my error. As described in Wikipedia, to take the limit of a function at a point $p\in S\subset\mathbb{R}$, it is required that $p\in\overline{S\setminus\{p\}}$. Unfortunately, in this case, $S=\{p\}$, so $p\not\in\overline{S\setminus\{p\}}=\emptyset$. So, although condition $(1)$ is satisfied, we cannot say that
$$\require{cancel}
\cancel{\lim_{x\to0}\sqrt{-x^2}=0}\tag2
$$
When $S=\mathbb{R}$, $p\in\overline{S\setminus\{p\}}$ for all $p\in S$, so $(1)$ is sufficient.
A: In complex analysis, the assertion $\lim_{z\to0}\sqrt{-z^2}=0$ is defensibly true even if you don't bother to specify the sign of the square root for nonzero numbers. But as a real-valued function of a real variable, the expression $\sqrt{-x^2}$ makes no sense except at $x=0$, so it really doesn't make sense to talk about its limit as $x$ tends to $0$. Most formal definitions of limit require the point being tended to to be a "limit point" for the domain of the function, meaning every (punctured) neighborhood of the point has nonempty intersection with the domain. The natural domain for the expression $\sqrt{-x^2}$ as a real-valued function of a real variable is the singleton set $\{0\}$, which has no limit points.
Added later: To put things another way, asking for the meaning of $\lim_{x\to c}f(x)$ when $c$ is not a limit point for the domain of $f$ is a little like asking for the meaning of $\gcd(\sqrt2,\sqrt3)$.  That is, the greatest common divisor is well defined for pairs of integers (not both $0$), but not for any old pair of numbers; likewise limits are only defined at limit points.  Expressions like $\gcd(\sqrt2,\sqrt3)$ and $\lim_{x\to0}\sqrt{-x^2}$ can, perhaps, be usefully likened to Noam Chomsky's grammatically correct but semantically nonsensical sentence "Colorless green ideas sleep furiously."
A: The best way to interpret this limit as a complex number. Which can be written as: $$\underset{x\to0}{lim}\hspace{0.3cm}x^2i+0$$
As always, you can separate both parts of the limit to get a more clear view of why the result is a real 0:
$$=\underset{x\to0}{lim}\hspace{0.3cm}x^2i+\underset{x\to0}{lim}\hspace{0.3cm}0 $$
And when you finally evaluate both parts of the limits, you can actually be sure that the real part and the imaginary part are both reaching 0 the closer x is to 0 from left or right (Even if the real part domain is $\{0\}$ since the imaginary part domain is $[0,\infty)$ ).
A: The function under consideration is $f:x \to \sqrt{-x^2} $ with domain {$0$}. In other words, your function is,  $f:$ {$0$} $\to ${$0$} with $f(0)=0.$ Your question is, "what is $\lim_{x \to 0}f(x)$?"
Now, let's talk about the domain. $0$ is not a limit point of {$0$} because every neighbourhood of $0$ in {$0$} is {$0$}, so there are no neighbourhoods of $0$ that contains a point $q \neq 0$ such that $q \in ${$0$}. Note that this is stronger than we needed: we only needed to show that there exists a neighbourhood of $0$ that does not contain a point $q \neq 0$ such that $q \in ${$0$}.
Now, in order for us to consider the limit $\ \lim_{x \to 0}f(x)$ (e.g. going by Rudin's definition), $\ 0$ must be a limit point of the domain which we have just seen it is not.
Therefore using the conventional definition of limit of a function at a point in the domain does not tell us anything about $\ \lim_{x \to 0}f(x),\ $ and in fact, I am not sure what the definition or meaning of "$\lim_{x \to 0}f(x)\ $" is for the above function, so let's stop talking about things that haven't been defined.
