Tomamos como el límite
$$\lim_{x \to 0^+}\left(\frac{- 3 x + \left(x^{2} + 2\right)}{2 x^{2} + \left(3 - 5 x\right)}\right)$$
cambiamos
$$\lim_{x \to 0^+}\left(\frac{- 3 x + \left(x^{2} + 2\right)}{2 x^{2} + \left(3 - 5 x\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\left(x - 2\right) \left(x - 1\right)}{\left(x - 1\right) \left(2 x - 3\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{x - 2}{2 x - 3}\right) = $$
$$\frac{-2}{-3 + 0 \cdot 2} = $$
= 2/3
Entonces la respuesta definitiva es:
$$\lim_{x \to 0^+}\left(\frac{- 3 x + \left(x^{2} + 2\right)}{2 x^{2} + \left(3 - 5 x\right)}\right) = \frac{2}{3}$$