Tomamos como el límite
$$\lim_{x \to 2^+}\left(\frac{- 5 x + \left(x^{2} + 6\right)}{x^{2} + 8}\right)$$
cambiamos
$$\lim_{x \to 2^+}\left(\frac{- 5 x + \left(x^{2} + 6\right)}{x^{2} + 8}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(x - 3\right) \left(x - 2\right)}{x^{2} + 8}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(x - 3\right) \left(x - 2\right)}{x^{2} + 8}\right) = $$
$$\frac{\left(-3 + 2\right) \left(-2 + 2\right)}{2^{2} + 8} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to 2^+}\left(\frac{- 5 x + \left(x^{2} + 6\right)}{x^{2} + 8}\right) = 0$$