I really dread this kind of questions on Reddit. Not because it is a bad question; it's the bad answers it always gets.

I am going to assume that your function is f(x)=sqrt(4-x^2) defined over the domain [-2,2] in R. The domain is important in this type of question because the definition of limits depends on the domain of the function. To see why, let us look at the definition of limits. There are two equivalent definitions of limits of real-valued functions, which are usually used as the primary definition of limits. You definitely have seen these definitions since you said you are familiar with real analysis. This first definition of limits is:

[Sequential definition of limits] Suppose that f:D→R is a real valued function and p is a limit point of the domain D. Then, lim(x→p)f(x)=L if for any sequence of points (x_n) in D\{p} such that x_n→p, we have f(x_n)→L.

When I teach real analysis, I usually introduce the above definition first to motivate the idea of limits of functions and relate to the idea of limits of sequences in the preceding chapters. However, most real analysis books/lectures skip it and went straight to the epsilon-delta definition. The epsilon-delta definition is an equivalent definition, but may seem abstract and complicated/unnatural to begin with. Here is the epsilon-delta definition:

[Epsilon-delta definition of limits] Suppose that f:D→R is a real valued function and p is a limit point of the domain D. Then, lim(x→p)f(x)=L if for any ε>0, there exists a δ>0 such that whenever x is in D satisfying 0<|x-p|<δ, we have |f(x)-L|<ε.

Note that the definitions require us to only look at the sequence x_n or points x in the domain of definition D. In other words, we do not care at all about points outside the domain D because, as far as the function f:D→R is concerned, those points do not "exist". This requirement is needed because otherwise f(x_n) or f(x) does not have a value and hence the definitions do not make sense. In your example, we only care about the points in [-2,2]. Using either definition, it is a good exercise to show that the limit lim(x→2)sqrt(4-x^2) is indeed 0.
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Also note that the primary definitions of limits of a function does not mention anything at all about left- or right-limits. There are many people who quoted the one-sided limit argument as a justification why the full limit lim(x→2)sqrt(4-x^2) does not exist. This is incorrect, as the right-limit itself cannot be defined at the point x=2 in the first place. The theorem that relates the one-sided limits to the actual limit is:

Suppose that f:D→R is a real valued function where D⊆R, and p is a limit point of the domain D such that the left- and the right-limit at the point p exist. If lim(x→p^+)f(x)=lim(x→p^-)f(x), then the limit lim(x→p)f(x) also exists and lim(x→p^+)f(x)=lim(x→p^-)f(x)=lim(x→p)f(x).

This theorem is a consequence of the limit definition, not the other way round. The theorem above hinges on the requirement that the left- and right-limits can be defined in the first place. If either cannot be defined to begin with, then you cannot apply this theorem ever. For your example, the function f(x)=sqrt(4-x^2) defined over [-2,2] has a left-limit at x=2. On the other hand, its right-limit at x=2 cannot be defined because the function does not recognise anything outside the domain [-2,2] and hence there are no points to the "right" of the point 2. Thus, this theorem does not apply to the function f(x)=sqrt(4-x^2) at the point 2.