Tomamos como el límite
$$\lim_{x \to 3^+}\left(\frac{- 6 x + \left(x^{2} + 8\right)}{x + 4}\right)$$
cambiamos
$$\lim_{x \to 3^+}\left(\frac{- 6 x + \left(x^{2} + 8\right)}{x + 4}\right)$$
=
$$\lim_{x \to 3^+}\left(\frac{\left(x - 4\right) \left(x - 2\right)}{x + 4}\right)$$
=
$$\lim_{x \to 3^+}\left(\frac{\left(x - 4\right) \left(x - 2\right)}{x + 4}\right) = $$
$$\frac{\left(-4 + 3\right) \left(-2 + 3\right)}{3 + 4} = $$
= -1/7
Entonces la respuesta definitiva es:
$$\lim_{x \to 3^+}\left(\frac{- 6 x + \left(x^{2} + 8\right)}{x + 4}\right) = - \frac{1}{7}$$