Tomamos como el límite
$$\lim_{x \to 3^+}\left(\frac{\left(x + 3\right) \left(x^{2} + \left(x + 1\right)\right)}{x^{3} + 27}\right)$$
cambiamos
$$\lim_{x \to 3^+}\left(\frac{\left(x + 3\right) \left(x^{2} + \left(x + 1\right)\right)}{x^{3} + 27}\right)$$
=
$$\lim_{x \to 3^+}\left(\frac{\left(x + 3\right) \left(x^{2} + x + 1\right)}{\left(x + 3\right) \left(x^{2} - 3 x + 9\right)}\right)$$
=
$$\lim_{x \to 3^+}\left(\frac{x^{2} + x + 1}{x^{2} - 3 x + 9}\right) = $$
$$\frac{1 + 3 + 3^{2}}{- 9 + 9 + 3^{2}} = $$
= 13/9
Entonces la respuesta definitiva es:
$$\lim_{x \to 3^+}\left(\frac{\left(x + 3\right) \left(x^{2} + \left(x + 1\right)\right)}{x^{3} + 27}\right) = \frac{13}{9}$$