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