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