Tomamos como el límite
$$\lim_{x \to 2^+}\left(\frac{x^{2} - 2 x}{6 x + \left(x^{2} - 4\right)}\right)$$
cambiamos
$$\lim_{x \to 2^+}\left(\frac{x^{2} - 2 x}{6 x + \left(x^{2} - 4\right)}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{x \left(x - 2\right)}{x^{2} + 6 x - 4}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{x \left(x - 2\right)}{x^{2} + 6 x - 4}\right) = $$
$$\frac{2 \left(-2 + 2\right)}{-4 + 2^{2} + 2 \cdot 6} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to 2^+}\left(\frac{x^{2} - 2 x}{6 x + \left(x^{2} - 4\right)}\right) = 0$$