Tomamos como el límite
$$\lim_{x \to 3^+}\left(\frac{- 4 x + \left(x^{2} + 3\right)}{2 x^{2} + \left(5 - 4 x\right)}\right)$$
cambiamos
$$\lim_{x \to 3^+}\left(\frac{- 4 x + \left(x^{2} + 3\right)}{2 x^{2} + \left(5 - 4 x\right)}\right)$$
=
$$\lim_{x \to 3^+}\left(\frac{\left(x - 3\right) \left(x - 1\right)}{2 x^{2} - 4 x + 5}\right)$$
=
$$\lim_{x \to 3^+}\left(\frac{\left(x - 3\right) \left(x - 1\right)}{2 x^{2} - 4 x + 5}\right) = $$
$$\frac{\left(-3 + 3\right) \left(-1 + 3\right)}{- 12 + 5 + 2 \cdot 3^{2}} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to 3^+}\left(\frac{- 4 x + \left(x^{2} + 3\right)}{2 x^{2} + \left(5 - 4 x\right)}\right) = 0$$