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