Tomamos como el límite
$$\lim_{x \to 0^+}\left(\frac{- 8 x + \left(x^{3} + 12\right)}{- 7 x + \left(x^{2} + 6\right)}\right)$$
cambiamos
$$\lim_{x \to 0^+}\left(\frac{- 8 x + \left(x^{3} + 12\right)}{- 7 x + \left(x^{2} + 6\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{x^{3} - 8 x + 12}{\left(x - 6\right) \left(x - 1\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{x^{3} - 8 x + 12}{\left(x - 6\right) \left(x - 1\right)}\right) = $$
$$\frac{0^{3} - 0 + 12}{\left(-6\right) \left(-1\right)} = $$
= 2
Entonces la respuesta definitiva es:
$$\lim_{x \to 0^+}\left(\frac{- 8 x + \left(x^{3} + 12\right)}{- 7 x + \left(x^{2} + 6\right)}\right) = 2$$