Tomamos como el límite
$$\lim_{x \to 2^+}\left(\frac{6 x + \left(3 x^{2} + \left(x^{3} - 12\right)\right)}{x - 4}\right)$$
cambiamos
$$\lim_{x \to 2^+}\left(\frac{6 x + \left(3 x^{2} + \left(x^{3} - 12\right)\right)}{x - 4}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{x^{3} + 3 x^{2} + 6 x - 12}{x - 4}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{x^{3} + 3 x^{2} + 6 x - 12}{x - 4}\right) = $$
$$\frac{-12 + 2^{3} + 3 \cdot 2^{2} + 2 \cdot 6}{-4 + 2} = $$
= -10
Entonces la respuesta definitiva es:
$$\lim_{x \to 2^+}\left(\frac{6 x + \left(3 x^{2} + \left(x^{3} - 12\right)\right)}{x - 4}\right) = -10$$