Tomamos como el límite
$$\lim_{x \to 1^+}\left(\frac{- x + \left(x^{2} - 2\right)}{2 x + \left(x^{2} - 8\right)}\right)$$
cambiamos
$$\lim_{x \to 1^+}\left(\frac{- x + \left(x^{2} - 2\right)}{2 x + \left(x^{2} - 8\right)}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{\left(x - 2\right) \left(x + 1\right)}{\left(x - 2\right) \left(x + 4\right)}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{x + 1}{x + 4}\right) = $$
$$\frac{1 + 1}{1 + 4} = $$
= 2/5
Entonces la respuesta definitiva es:
$$\lim_{x \to 1^+}\left(\frac{- x + \left(x^{2} - 2\right)}{2 x + \left(x^{2} - 8\right)}\right) = \frac{2}{5}$$