Tomamos como el límite
$$\lim_{x \to 0^+}\left(\frac{- 6 x + \left(x^{2} + 8\right)}{- 8 x + \left(x^{2} + 12\right)}\right)$$
cambiamos
$$\lim_{x \to 0^+}\left(\frac{- 6 x + \left(x^{2} + 8\right)}{- 8 x + \left(x^{2} + 12\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\left(x - 4\right) \left(x - 2\right)}{\left(x - 6\right) \left(x - 2\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{x - 4}{x - 6}\right) = $$
$$\frac{-4}{-6} = $$
= 2/3
Entonces la respuesta definitiva es:
$$\lim_{x \to 0^+}\left(\frac{- 6 x + \left(x^{2} + 8\right)}{- 8 x + \left(x^{2} + 12\right)}\right) = \frac{2}{3}$$