Tomamos como el límite
$$\lim_{x \to 2^+}\left(\frac{3 x^{2} + \left(- 2 x - 1\right)}{x + 4}\right)$$
cambiamos
$$\lim_{x \to 2^+}\left(\frac{3 x^{2} + \left(- 2 x - 1\right)}{x + 4}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(x - 1\right) \left(3 x + 1\right)}{x + 4}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(x - 1\right) \left(3 x + 1\right)}{x + 4}\right) = $$
$$\frac{\left(-1 + 2\right) \left(1 + 2 \cdot 3\right)}{2 + 4} = $$
= 7/6
Entonces la respuesta definitiva es:
$$\lim_{x \to 2^+}\left(\frac{3 x^{2} + \left(- 2 x - 1\right)}{x + 4}\right) = \frac{7}{6}$$