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