Tomamos como el límite
$$\lim_{x \to 0^+}\left(\frac{5 x + \left(x^{2} + 6\right)}{- 3 x + \left(x^{2} + 2\right)}\right)$$
cambiamos
$$\lim_{x \to 0^+}\left(\frac{5 x + \left(x^{2} + 6\right)}{- 3 x + \left(x^{2} + 2\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\left(x + 2\right) \left(x + 3\right)}{\left(x - 2\right) \left(x - 1\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\left(x + 2\right) \left(x + 3\right)}{\left(x - 2\right) \left(x - 1\right)}\right) = $$
$$\frac{2 \cdot 3}{\left(-2\right) \left(-1\right)} = $$
= 3
Entonces la respuesta definitiva es:
$$\lim_{x \to 0^+}\left(\frac{5 x + \left(x^{2} + 6\right)}{- 3 x + \left(x^{2} + 2\right)}\right) = 3$$