so... checked it using L'Hopital rule... and it just went for a cycle of 0/0.
so instead, let's check it by the independent limits above and below.
[tex]\bf \lim\limits_{x\to 2^-}~\cfrac{\sqrt{4-x^2}}{x-2}\implies \cfrac{\lim\limits_{x\to 2^-}~\sqrt{4-x^2}}{\lim\limits_{x\to 2^-}~x-2}[/tex]
now, if you look at the table of values for "x", check picture below, starting at -10 and ever approaching 2 from the left, notice that the numerator is decreasing, but slowly on every value for "x", whilst the denominator is decreasing as well, but faster than the numerator, with slightly larger jumps.
After they pass x = 0, you can see how tiny the denominator is turning, and the numerator is getting a bit larger than the denominator. or as the lingo goes, "the denominator is moving faster than the numerator".
because that will yield an ever tinier denominator than the numerator, the fraction becomes a larger value, negative all the while, because the denominator is negative while the numerator is positive.
since the fraction is then become negatively larger and larger, the one-sided limit will then be -∞.
