Tomamos como el límite
$$\lim_{x \to 4^+}\left(\frac{x^{2} - 16}{6 x + \left(x^{2} + 8\right)}\right)$$
cambiamos
$$\lim_{x \to 4^+}\left(\frac{x^{2} - 16}{6 x + \left(x^{2} + 8\right)}\right)$$
=
$$\lim_{x \to 4^+}\left(\frac{\left(x - 4\right) \left(x + 4\right)}{\left(x + 2\right) \left(x + 4\right)}\right)$$
=
$$\lim_{x \to 4^+}\left(\frac{x - 4}{x + 2}\right) = $$
$$\frac{-4 + 4}{2 + 4} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to 4^+}\left(\frac{x^{2} - 16}{6 x + \left(x^{2} + 8\right)}\right) = 0$$