Tomamos como el límite
$$\lim_{x \to -1^+}\left(\frac{- 4 x + \left(x^{2} - 5\right)}{- 5 x + \left(x^{3} + 2\right)}\right)$$
cambiamos
$$\lim_{x \to -1^+}\left(\frac{- 4 x + \left(x^{2} - 5\right)}{- 5 x + \left(x^{3} + 2\right)}\right)$$
=
$$\lim_{x \to -1^+}\left(\frac{\left(x - 5\right) \left(x + 1\right)}{\left(x - 2\right) \left(x^{2} + 2 x - 1\right)}\right)$$
=
$$\lim_{x \to -1^+}\left(\frac{x^{2} - 4 x - 5}{x^{3} - 5 x + 2}\right) = $$
$$\frac{-5 + \left(-1\right)^{2} - -4}{\left(-1\right)^{3} + 2 - -5} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to -1^+}\left(\frac{- 4 x + \left(x^{2} - 5\right)}{- 5 x + \left(x^{3} + 2\right)}\right) = 0$$