Tomamos como el límite
$$\lim_{x \to 3^+}\left(\frac{12 x + \left(x^{2} - 4\right)}{x^{3} + 6 x}\right)$$
cambiamos
$$\lim_{x \to 3^+}\left(\frac{12 x + \left(x^{2} - 4\right)}{x^{3} + 6 x}\right)$$
=
$$\lim_{x \to 3^+}\left(\frac{x^{2} + 12 x - 4}{x \left(x^{2} + 6\right)}\right)$$
=
$$\lim_{x \to 3^+}\left(\frac{x^{2} + 12 x - 4}{x \left(x^{2} + 6\right)}\right) = $$
$$\frac{-4 + 3^{2} + 3 \cdot 12}{3 \left(6 + 3^{2}\right)} = $$
= 41/45
Entonces la respuesta definitiva es:
$$\lim_{x \to 3^+}\left(\frac{12 x + \left(x^{2} - 4\right)}{x^{3} + 6 x}\right) = \frac{41}{45}$$