Tomamos como el límite
$$\lim_{x \to 3^+}\left(\frac{x^{2} + \left(x - 12\right)}{x^{2} - 13}\right)$$
cambiamos
$$\lim_{x \to 3^+}\left(\frac{x^{2} + \left(x - 12\right)}{x^{2} - 13}\right)$$
=
$$\lim_{x \to 3^+}\left(\frac{\left(x - 3\right) \left(x + 4\right)}{x^{2} - 13}\right)$$
=
$$\lim_{x \to 3^+}\left(\frac{\left(x - 3\right) \left(x + 4\right)}{x^{2} - 13}\right) = $$
$$\frac{\left(-3 + 3\right) \left(3 + 4\right)}{-13 + 3^{2}} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to 3^+}\left(\frac{x^{2} + \left(x - 12\right)}{x^{2} - 13}\right) = 0$$