Tomamos como el límite
$$\lim_{x \to 4^+}\left(\frac{x^{2} - 16}{- 3 x + \left(x^{2} + 4\right)}\right)$$
cambiamos
$$\lim_{x \to 4^+}\left(\frac{x^{2} - 16}{- 3 x + \left(x^{2} + 4\right)}\right)$$
=
$$\lim_{x \to 4^+}\left(\frac{\left(x - 4\right) \left(x + 4\right)}{x^{2} - 3 x + 4}\right)$$
=
$$\lim_{x \to 4^+}\left(\frac{x^{2} - 16}{x^{2} - 3 x + 4}\right) = $$
$$\frac{-16 + 4^{2}}{- 12 + 4 + 4^{2}} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to 4^+}\left(\frac{x^{2} - 16}{- 3 x + \left(x^{2} + 4\right)}\right) = 0$$