Tomamos como el límite
$$\lim_{x \to -2^+}\left(\frac{x^{2} + \left(x - 2\right)}{2 x + \left(6 - 3 x^{2}\right)}\right)$$
cambiamos
$$\lim_{x \to -2^+}\left(\frac{x^{2} + \left(x - 2\right)}{2 x + \left(6 - 3 x^{2}\right)}\right)$$
=
$$\lim_{x \to -2^+}\left(\frac{\left(x - 1\right) \left(x + 2\right)}{- 3 x^{2} + 2 x + 6}\right)$$
=
$$\lim_{x \to -2^+}\left(\frac{\left(x - 1\right) \left(x + 2\right)}{- 3 x^{2} + 2 x + 6}\right) = $$
$$\frac{\left(-2 - 1\right) \left(-2 + 2\right)}{- 3 \left(-2\right)^{2} + \left(-2\right) 2 + 6} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to -2^+}\left(\frac{x^{2} + \left(x - 2\right)}{2 x + \left(6 - 3 x^{2}\right)}\right) = 0$$