Tomamos como el límite
$$\lim_{x \to 0^+}\left(\frac{4 x + \left(x^{2} + 1\right)}{2 x^{3} + \left(x^{2} + 5\right)}\right)$$
cambiamos
$$\lim_{x \to 0^+}\left(\frac{4 x + \left(x^{2} + 1\right)}{2 x^{3} + \left(x^{2} + 5\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{x^{2} + 4 x + 1}{2 x^{3} + x^{2} + 5}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{x^{2} + 4 x + 1}{2 x^{3} + x^{2} + 5}\right) = $$
$$\frac{0^{2} + 0 \cdot 4 + 1}{0^{2} + 2 \cdot 0^{3} + 5} = $$
= 1/5
Entonces la respuesta definitiva es:
$$\lim_{x \to 0^+}\left(\frac{4 x + \left(x^{2} + 1\right)}{2 x^{3} + \left(x^{2} + 5\right)}\right) = \frac{1}{5}$$