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