Tomamos como el límite
$$\lim_{x \to 2^+}\left(\frac{x \left(x - 2\right) \left(- 5 x + \left(x^{2} + 6\right)\right)}{x + 4}\right)$$
cambiamos
$$\lim_{x \to 2^+}\left(\frac{x \left(x - 2\right) \left(- 5 x + \left(x^{2} + 6\right)\right)}{x + 4}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{x \left(x - 3\right) \left(x - 2\right)^{2}}{x + 4}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{x \left(x - 3\right) \left(x - 2\right)^{2}}{x + 4}\right) = $$
$$\frac{2 \left(-3 + 2\right) \left(-2 + 2\right)^{2}}{2 + 4} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to 2^+}\left(\frac{x \left(x - 2\right) \left(- 5 x + \left(x^{2} + 6\right)\right)}{x + 4}\right) = 0$$