Tomamos como el límite
$$\lim_{x \to 4^+}\left(\frac{6 x + \left(x^{3} - 8\right)}{- 3 x^{2} + \left(x^{3} + 4\right)}\right)$$
cambiamos
$$\lim_{x \to 4^+}\left(\frac{6 x + \left(x^{3} - 8\right)}{- 3 x^{2} + \left(x^{3} + 4\right)}\right)$$
=
$$\lim_{x \to 4^+}\left(\frac{x^{3} + 6 x - 8}{\left(x - 2\right)^{2} \left(x + 1\right)}\right)$$
=
$$\lim_{x \to 4^+}\left(\frac{x^{3} + 6 x - 8}{\left(x - 2\right)^{2} \left(x + 1\right)}\right) = $$
$$\frac{-8 + 4 \cdot 6 + 4^{3}}{\left(-2 + 4\right)^{2} \left(1 + 4\right)} = $$
= 4
Entonces la respuesta definitiva es:
$$\lim_{x \to 4^+}\left(\frac{6 x + \left(x^{3} - 8\right)}{- 3 x^{2} + \left(x^{3} + 4\right)}\right) = 4$$