Tomamos como el límite
$$\lim_{x \to 2^+}\left(\frac{13 x + \left(5 x^{2} + 6\right)}{3 x^{3} - 8}\right)$$
cambiamos
$$\lim_{x \to 2^+}\left(\frac{13 x + \left(5 x^{2} + 6\right)}{3 x^{3} - 8}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(x + 2\right) \left(5 x + 3\right)}{3 x^{3} - 8}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(x + 2\right) \left(5 x + 3\right)}{3 x^{3} - 8}\right) = $$
$$\frac{\left(2 + 2\right) \left(3 + 2 \cdot 5\right)}{-8 + 3 \cdot 2^{3}} = $$
= 13/4
Entonces la respuesta definitiva es:
$$\lim_{x \to 2^+}\left(\frac{13 x + \left(5 x^{2} + 6\right)}{3 x^{3} - 8}\right) = \frac{13}{4}$$