Tomamos como el límite
$$\lim_{x \to x_{0}^+}\left(\frac{2 x^{2} + \left(x - 1\right)}{- 3 x + \left(x^{2} - 4\right)}\right)$$
cambiamos
$$\lim_{x \to x_{0}^+}\left(\frac{2 x^{2} + \left(x - 1\right)}{- 3 x + \left(x^{2} - 4\right)}\right)$$
=
$$\lim_{x \to x_{0}^+}\left(\frac{\left(x + 1\right) \left(2 x - 1\right)}{\left(x - 4\right) \left(x + 1\right)}\right)$$
=
$$\lim_{x \to x_{0}^+}\left(\frac{2 x - 1}{x - 4}\right) = $$
$$\frac{2 x_{0} - 1}{x_{0} - 4} = $$
= (-1 + 2*x0)/(-4 + x0)
Entonces la respuesta definitiva es:
$$\lim_{x \to x_{0}^+}\left(\frac{2 x^{2} + \left(x - 1\right)}{- 3 x + \left(x^{2} - 4\right)}\right) = \frac{2 x_{0} - 1}{x_{0} - 4}$$