Tomamos como el límite
$$\lim_{x \to 3^+}\left(\frac{x^{3} - 27}{3 x + \left(x^{2} + 18\right)}\right)$$
cambiamos
$$\lim_{x \to 3^+}\left(\frac{x^{3} - 27}{3 x + \left(x^{2} + 18\right)}\right)$$
=
$$\lim_{x \to 3^+}\left(\frac{\left(x - 3\right) \left(x^{2} + 3 x + 9\right)}{x^{2} + 3 x + 18}\right)$$
=
$$\lim_{x \to 3^+}\left(\frac{x^{3} - 27}{x^{2} + 3 x + 18}\right) = $$
$$\frac{-27 + 3^{3}}{3^{2} + 3 \cdot 3 + 18} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to 3^+}\left(\frac{x^{3} - 27}{3 x + \left(x^{2} + 18\right)}\right) = 0$$