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