Tomamos como el límite
$$\lim_{x \to 2^+}\left(\frac{- x + \left(x^{2} - 2\right)}{2 x + \left(x^{2} + 8\right)}\right)$$
cambiamos
$$\lim_{x \to 2^+}\left(\frac{- x + \left(x^{2} - 2\right)}{2 x + \left(x^{2} + 8\right)}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(x - 2\right) \left(x + 1\right)}{x^{2} + 2 x + 8}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(x - 2\right) \left(x + 1\right)}{x^{2} + 2 x + 8}\right) = $$
$$\frac{\left(-2 + 2\right) \left(1 + 2\right)}{2^{2} + 2 \cdot 2 + 8} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to 2^+}\left(\frac{- x + \left(x^{2} - 2\right)}{2 x + \left(x^{2} + 8\right)}\right) = 0$$