Tomamos como el límite
$$\lim_{x \to 2^+}\left(\frac{8 - 2 x^{2}}{4 x + \left(x^{2} + 12\right)}\right)$$
cambiamos
$$\lim_{x \to 2^+}\left(\frac{8 - 2 x^{2}}{4 x + \left(x^{2} + 12\right)}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(-1\right) 2 \left(x - 2\right) \left(x + 2\right)}{x^{2} + 4 x + 12}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{8 - 2 x^{2}}{x^{2} + 4 x + 12}\right) = $$
$$\frac{8 - 2 \cdot 2^{2}}{2^{2} + 2 \cdot 4 + 12} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to 2^+}\left(\frac{8 - 2 x^{2}}{4 x + \left(x^{2} + 12\right)}\right) = 0$$