Tomamos como el límite
$$\lim_{x \to 4^+}\left(\frac{x^{2} + \left(x - 12\right)}{2 x + \left(x^{2} - 8\right)}\right)$$
cambiamos
$$\lim_{x \to 4^+}\left(\frac{x^{2} + \left(x - 12\right)}{2 x + \left(x^{2} - 8\right)}\right)$$
=
$$\lim_{x \to 4^+}\left(\frac{\left(x - 3\right) \left(x + 4\right)}{\left(x - 2\right) \left(x + 4\right)}\right)$$
=
$$\lim_{x \to 4^+}\left(\frac{x - 3}{x - 2}\right) = $$
$$\frac{-3 + 4}{-2 + 4} = $$
= 1/2
Entonces la respuesta definitiva es:
$$\lim_{x \to 4^+}\left(\frac{x^{2} + \left(x - 12\right)}{2 x + \left(x^{2} - 8\right)}\right) = \frac{1}{2}$$