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