Tomamos como el límite
$$\lim_{x \to 2^+}\left(\frac{- 4 x^{2} + \left(x^{3} + 8\right)}{x^{4} - 1}\right)$$
cambiamos
$$\lim_{x \to 2^+}\left(\frac{- 4 x^{2} + \left(x^{3} + 8\right)}{x^{4} - 1}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(x - 2\right) \left(x^{2} - 2 x - 4\right)}{\left(x - 1\right) \left(x + 1\right) \left(x^{2} + 1\right)}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{x^{3} - 4 x^{2} + 8}{x^{4} - 1}\right) = $$
$$\frac{- 4 \cdot 2^{2} + 8 + 2^{3}}{-1 + 2^{4}} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to 2^+}\left(\frac{- 4 x^{2} + \left(x^{3} + 8\right)}{x^{4} - 1}\right) = 0$$