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