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