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