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