Tomamos como el límite
$$\lim_{x \to -2^+}\left(\frac{4 x^{3} + \left(2 x^{2} + 1\right)}{3 x^{3} - 5}\right)$$
cambiamos
$$\lim_{x \to -2^+}\left(\frac{4 x^{3} + \left(2 x^{2} + 1\right)}{3 x^{3} - 5}\right)$$
=
$$\lim_{x \to -2^+}\left(\frac{4 x^{3} + 2 x^{2} + 1}{3 x^{3} - 5}\right)$$
=
$$\lim_{x \to -2^+}\left(\frac{4 x^{3} + 2 x^{2} + 1}{3 x^{3} - 5}\right) = $$
$$\frac{4 \left(-2\right)^{3} + 1 + 2 \left(-2\right)^{2}}{3 \left(-2\right)^{3} - 5} = $$
= 23/29
Entonces la respuesta definitiva es:
$$\lim_{x \to -2^+}\left(\frac{4 x^{3} + \left(2 x^{2} + 1\right)}{3 x^{3} - 5}\right) = \frac{23}{29}$$