Tomamos como el límite
$$\lim_{x \to -1^+}\left(\frac{12 x + \left(8 - x^{2}\right)}{4 x + 5}\right)$$
cambiamos
$$\lim_{x \to -1^+}\left(\frac{12 x + \left(8 - x^{2}\right)}{4 x + 5}\right)$$
=
$$\lim_{x \to -1^+}\left(\frac{- x^{2} + 12 x + 8}{4 x + 5}\right)$$
=
$$\lim_{x \to -1^+}\left(\frac{- x^{2} + 12 x + 8}{4 x + 5}\right) = $$
$$\frac{\left(-1\right) 12 - \left(-1\right)^{2} + 8}{\left(-1\right) 4 + 5} = $$
= -5
Entonces la respuesta definitiva es:
$$\lim_{x \to -1^+}\left(\frac{12 x + \left(8 - x^{2}\right)}{4 x + 5}\right) = -5$$