Tomamos como el límite
$$\lim_{x \to 2^+}\left(\frac{8 x + \left(x^{2} - 9\right)}{6 x + \left(x^{2} - 27\right)}\right)$$
cambiamos
$$\lim_{x \to 2^+}\left(\frac{8 x + \left(x^{2} - 9\right)}{6 x + \left(x^{2} - 27\right)}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(x - 1\right) \left(x + 9\right)}{\left(x - 3\right) \left(x + 9\right)}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{x - 1}{x - 3}\right) = $$
$$\frac{-1 + 2}{-3 + 2} = $$
= -1
Entonces la respuesta definitiva es:
$$\lim_{x \to 2^+}\left(\frac{8 x + \left(x^{2} - 9\right)}{6 x + \left(x^{2} - 27\right)}\right) = -1$$