Tomamos como el límite
$$\lim_{x \to 2^+}\left(\frac{5 x + \left(2 x^{2} - 3\right)}{5 x + \left(x^{2} + 6\right)}\right)$$
cambiamos
$$\lim_{x \to 2^+}\left(\frac{5 x + \left(2 x^{2} - 3\right)}{5 x + \left(x^{2} + 6\right)}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(x + 3\right) \left(2 x - 1\right)}{\left(x + 2\right) \left(x + 3\right)}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{2 x - 1}{x + 2}\right) = $$
$$\frac{-1 + 2 \cdot 2}{2 + 2} = $$
= 3/4
Entonces la respuesta definitiva es:
$$\lim_{x \to 2^+}\left(\frac{5 x + \left(2 x^{2} - 3\right)}{5 x + \left(x^{2} + 6\right)}\right) = \frac{3}{4}$$