Tomamos como el límite
$$\lim_{x \to 3^+}\left(\frac{- 2 x + \left(x^{2} - 3\right)}{4 x + 12}\right)$$
cambiamos
$$\lim_{x \to 3^+}\left(\frac{- 2 x + \left(x^{2} - 3\right)}{4 x + 12}\right)$$
=
$$\lim_{x \to 3^+}\left(\frac{\left(x - 3\right) \left(x + 1\right)}{4 x + 12}\right)$$
=
$$\lim_{x \to 3^+}\left(\frac{\left(x - 3\right) \left(x + 1\right)}{4 \left(x + 3\right)}\right) = $$
$$\frac{\left(-3 + 3\right) \left(1 + 3\right)}{4 \left(3 + 3\right)} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to 3^+}\left(\frac{- 2 x + \left(x^{2} - 3\right)}{4 x + 12}\right) = 0$$