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