Tomamos como el límite
$$\lim_{x \to -4^+}\left(\frac{5 x + \left(x^{2} + 6\right)}{x^{2} + \left(x - 11\right)}\right)$$
cambiamos
$$\lim_{x \to -4^+}\left(\frac{5 x + \left(x^{2} + 6\right)}{x^{2} + \left(x - 11\right)}\right)$$
=
$$\lim_{x \to -4^+}\left(\frac{\left(x + 2\right) \left(x + 3\right)}{x^{2} + x - 11}\right)$$
=
$$\lim_{x \to -4^+}\left(\frac{\left(x + 2\right) \left(x + 3\right)}{x^{2} + x - 11}\right) = $$
$$\frac{\left(-4 + 2\right) \left(-4 + 3\right)}{-11 - 4 + \left(-4\right)^{2}} = $$
= 2
Entonces la respuesta definitiva es:
$$\lim_{x \to -4^+}\left(\frac{5 x + \left(x^{2} + 6\right)}{x^{2} + \left(x - 11\right)}\right) = 2$$