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