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