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