Tomamos como el límite
$$\lim_{x \to 2^+}\left(\frac{4 x^{2} + \left(1 - 5 x\right)}{2 x^{2} + \left(1 - x\right)}\right)$$
cambiamos
$$\lim_{x \to 2^+}\left(\frac{4 x^{2} + \left(1 - 5 x\right)}{2 x^{2} + \left(1 - x\right)}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(x - 1\right) \left(4 x - 1\right)}{2 x^{2} - x + 1}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(x - 1\right) \left(4 x - 1\right)}{2 x^{2} - x + 1}\right) = $$
$$\frac{\left(-1 + 2\right) \left(-1 + 2 \cdot 4\right)}{- 2 + 1 + 2 \cdot 2^{2}} = $$
= 1
Entonces la respuesta definitiva es:
$$\lim_{x \to 2^+}\left(\frac{4 x^{2} + \left(1 - 5 x\right)}{2 x^{2} + \left(1 - x\right)}\right) = 1$$