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