$ \displaystyle\lim_{x \to 3^-} [log_{\frac{1}{2}} (9-x^2)] = $
$ \displaystyle\lim_{x \to 3^-} [log_{\frac{1}{2}} (3-x) + log_{\frac{1}{2}} (3+x) ] = $
poniamo t = 3 - x ⇒ x = 3 - t inoltre se x → 3⁻ allora t → 0⁺
$ \displaystyle\lim_{t \to 0^+} [log_{\frac{1}{2}} (t) + log_{\frac{1}{2}} (3+3-t) ] = $
$ = +\infty + log_{\frac{1}{2}} (6) = +\infty $