Tomamos como el límite
$$\lim_{x \to 2^+}\left(\frac{2 x + \left(x^{2} - 8\right)}{6 - 5 x}\right)$$
cambiamos
$$\lim_{x \to 2^+}\left(\frac{2 x + \left(x^{2} - 8\right)}{6 - 5 x}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(x - 2\right) \left(x + 4\right)}{6 - 5 x}\right)$$
=
$$\lim_{x \to 2^+}\left(- \frac{\left(x - 2\right) \left(x + 4\right)}{5 x - 6}\right) = $$
$$- \frac{\left(-2 + 2\right) \left(2 + 4\right)}{-6 + 2 \cdot 5} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to 2^+}\left(\frac{2 x + \left(x^{2} - 8\right)}{6 - 5 x}\right) = 0$$