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