Tomamos como el límite
$$\lim_{x \to 2^+}\left(\frac{3 x^{2} + 8}{x^{2} + \left(x - 2\right)}\right)$$
cambiamos
$$\lim_{x \to 2^+}\left(\frac{3 x^{2} + 8}{x^{2} + \left(x - 2\right)}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{3 x^{2} + 8}{\left(x - 1\right) \left(x + 2\right)}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{3 x^{2} + 8}{\left(x - 1\right) \left(x + 2\right)}\right) = $$
$$\frac{8 + 3 \cdot 2^{2}}{\left(-1 + 2\right) \left(2 + 2\right)} = $$
= 5
Entonces la respuesta definitiva es:
$$\lim_{x \to 2^+}\left(\frac{3 x^{2} + 8}{x^{2} + \left(x - 2\right)}\right) = 5$$