Tomamos como el límite
$$\lim_{x \to 1^+}\left(\frac{- x + \left(x^{2} - 2\right)}{2 x^{2} + \left(x - 3\right)}\right)$$
cambiamos
$$\lim_{x \to 1^+}\left(\frac{- x + \left(x^{2} - 2\right)}{2 x^{2} + \left(x - 3\right)}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{\left(x - 2\right) \left(x + 1\right)}{\left(x - 1\right) \left(2 x + 3\right)}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{\left(x - 2\right) \left(x + 1\right)}{\left(x - 1\right) \left(2 x + 3\right)}\right) = $$
False
= -oo
Entonces la respuesta definitiva es:
$$\lim_{x \to 1^+}\left(\frac{- x + \left(x^{2} - 2\right)}{2 x^{2} + \left(x - 3\right)}\right) = -\infty$$