Tomamos como el límite
$$\lim_{x \to -4^+}\left(\frac{5 x + 6}{- 4 x + \left(x^{2} + 3\right)}\right)$$
cambiamos
$$\lim_{x \to -4^+}\left(\frac{5 x + 6}{- 4 x + \left(x^{2} + 3\right)}\right)$$
=
$$\lim_{x \to -4^+}\left(\frac{5 x + 6}{\left(x - 3\right) \left(x - 1\right)}\right)$$
=
$$\lim_{x \to -4^+}\left(\frac{5 x + 6}{\left(x - 3\right) \left(x - 1\right)}\right) = $$
$$\frac{\left(-4\right) 5 + 6}{\left(-4 - 3\right) \left(-4 - 1\right)} = $$
= -2/5
Entonces la respuesta definitiva es:
$$\lim_{x \to -4^+}\left(\frac{5 x + 6}{- 4 x + \left(x^{2} + 3\right)}\right) = - \frac{2}{5}$$