How do you find the limit of #sqrt((4-x^2))# as x approaches #-2#?

The domain of #f(x) = sqrt(4-x^2)# is #[-2,2]#

There is no limit as #x# approaches #-2# from the left, because the function does not give real values for such #x#.
We do see that as #x# approaches #-2# from the right, we get values of #f(x)# approaching #0#.

If your definition has not provided for this kind of thing, we must say

#lim_(xrarr-2)sqrt(4-x^2)# #" "# does not exist.

For example, the definition in Stewart's Calculus is

Let #f# be defined on some open interval containing #c# except possible at #c# itself. Then #lim_(xrarrc)f(x) = L# if and only if

for every positive #epsilon#, there is a positive #delta# such that for all #x#, we have #0 < abs(x-c) < delta# implies #abs(f(x)-L) < epsilon#.

#f(x) = sqrt(4-x^2)# is not defined (real valued) on an open interval containing #-2#. So the limit as #x# approaches #-2# is not defined.

Note also, that in this definition, the conditional must hold for all #x# on either side of #c#.

Some treatments of the limit include in the definition a phrase that tells us to only consider values in the domain of #f# when evaluating limits.
They might say, for example
Let #f# be defined for some real numbers in an open interval containing #c#. Then #lim_(xrarrc)f(x) = L# if and only if;

for every positive #epsilon#, there is a positive #delta# such that for all #x# in the domain of #f#, we have #0 < abs(x-c) < delta# implies #abs(f(x)-L) < epsilon#.

For such a definition, we get

#lim_(xrarr-2)sqrt(4-x^2) = 0#