Tomamos como el límite
$$\lim_{x \to 3^+}\left(\frac{3 x^{2} + \left(2 x - 16\right)}{- x + \left(x^{2} - 2\right)}\right)$$
cambiamos
$$\lim_{x \to 3^+}\left(\frac{3 x^{2} + \left(2 x - 16\right)}{- x + \left(x^{2} - 2\right)}\right)$$
=
$$\lim_{x \to 3^+}\left(\frac{\left(x - 2\right) \left(3 x + 8\right)}{\left(x - 2\right) \left(x + 1\right)}\right)$$
=
$$\lim_{x \to 3^+}\left(\frac{3 x + 8}{x + 1}\right) = $$
$$\frac{8 + 3 \cdot 3}{1 + 3} = $$
= 17/4
Entonces la respuesta definitiva es:
$$\lim_{x \to 3^+}\left(\frac{3 x^{2} + \left(2 x - 16\right)}{- x + \left(x^{2} - 2\right)}\right) = \frac{17}{4}$$