Tomamos como el límite
$$\lim_{x \to -2^+}\left(\frac{- 2 x + \left(x^{2} - 8\right)}{x^{3} - 8}\right)$$
cambiamos
$$\lim_{x \to -2^+}\left(\frac{- 2 x + \left(x^{2} - 8\right)}{x^{3} - 8}\right)$$
=
$$\lim_{x \to -2^+}\left(\frac{\left(x - 4\right) \left(x + 2\right)}{\left(x - 2\right) \left(x^{2} + 2 x + 4\right)}\right)$$
=
$$\lim_{x \to -2^+}\left(\frac{x^{2} - 2 x - 8}{x^{3} - 8}\right) = $$
$$\frac{-8 + \left(-2\right)^{2} - -4}{-8 + \left(-2\right)^{3}} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to -2^+}\left(\frac{- 2 x + \left(x^{2} - 8\right)}{x^{3} - 8}\right) = 0$$