Tomamos como el límite
$$\lim_{x \to -4^+}\left(\frac{x^{2} - 16}{12 x + \left(x^{2} - 32\right)}\right)$$
cambiamos
$$\lim_{x \to -4^+}\left(\frac{x^{2} - 16}{12 x + \left(x^{2} - 32\right)}\right)$$
=
$$\lim_{x \to -4^+}\left(\frac{\left(x - 4\right) \left(x + 4\right)}{x^{2} + 12 x - 32}\right)$$
=
$$\lim_{x \to -4^+}\left(\frac{x^{2} - 16}{x^{2} + 12 x - 32}\right) = $$
$$\frac{-16 + \left(-4\right)^{2}}{\left(-4\right) 12 - 32 + \left(-4\right)^{2}} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to -4^+}\left(\frac{x^{2} - 16}{12 x + \left(x^{2} - 32\right)}\right) = 0$$