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