The function itself is not well-defined at [tex]x=2[/tex] because both numerator and denominator are equal to zero there, so it's discontinuous there. We attempt to remove this discontinuity.
Rationalize the numerator and denominator.
[tex]\displaystyle \frac{x - \sqrt{8 - x^2}}{\sqrt{x^2 + 12} - 4} \cdot \frac{x + \sqrt{8 - x^2}}{x + \sqrt{8 - x^2}} \cdot \frac{\sqrt{x^2+12} + 4}{\sqrt{x^2 + 12} + 4}[/tex]
[tex]\displaystyle = \frac{x^2 - \left(\sqrt{8-x^2}\right)^2}{\left(\sqrt{x^2+12}\right)^2 - 4^2} \cdot \frac{\sqrt{x^2 + 12} + 4}{x + \sqrt{8 - x^2}}[/tex]
[tex]\displaystyle = \frac{x^2 - (8-x^2)}{(x^2+12)-16} \cdot \frac{\sqrt{x^2 + 12} + 4}{x + \sqrt{8 - x^2}}[/tex]
[tex]\displaystyle = \frac{2x^2 - 8}{x^2 - 4} \cdot \frac{\sqrt{x^2 + 12} + 4}{x + \sqrt{8 - x^2}}[/tex]
[tex]\displaystyle = 2 \frac{(x-2)(x+2)}{(x-2)(x+4)} \cdot \frac{\sqrt{x^2 + 12} + 4}{x + \sqrt{8 - x^2}}[/tex]
Then in the limit, since [tex]x\neq2[/tex], we can cancel both factors of [tex]x-2[/tex] and [tex]x+2[/tex], subsequently removing the discontinuities at [tex]x=\pm2[/tex], and we're left with
[tex]\displaystyle \lim_{x\to2} \frac{x-\sqrt{8-x^2}}{\sqrt{x^2+12}-4} = 2 \lim_{x\to2} \frac{\sqrt{x^2+12}+4}{x+\sqrt{8-x^2}}[/tex]
which is now continuous at [tex]x=2[/tex], so we can evaluate directly.
[tex]\displaystyle \lim_{x\to2} \frac{x-\sqrt{8-x^2}}{\sqrt{x^2+12}-4} = 2 \cdot \frac{\sqrt{2^2+12}+4}{2+\sqrt{8-2^2}} = \frac{16}4 = \boxed{4}[/tex]
