Tomamos como el límite
$$\lim_{x \to 2^+}\left(\frac{\frac{x^{2}}{2} - 2}{\frac{2 x}{3} - 3}\right)$$
cambiamos
$$\lim_{x \to 2^+}\left(\frac{\frac{x^{2}}{2} - 2}{\frac{2 x}{3} - 3}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\frac{1}{2} \left(x - 2\right) \left(x + 2\right)}{\frac{2 x}{3} - 3}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{3 \left(x^{2} - 4\right)}{2 \left(2 x - 9\right)}\right) = $$
$$\frac{3 \left(-4 + 2^{2}\right)}{2 \left(-9 + 2 \cdot 2\right)} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to 2^+}\left(\frac{\frac{x^{2}}{2} - 2}{\frac{2 x}{3} - 3}\right) = 0$$