Tomamos como el límite
$$\lim_{x \to 1^+}\left(\frac{4 x + \left(- 7 x^{2} + \left(x^{3} + 2\right)\right)}{2 x - 5}\right)$$
cambiamos
$$\lim_{x \to 1^+}\left(\frac{4 x + \left(- 7 x^{2} + \left(x^{3} + 2\right)\right)}{2 x - 5}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{\left(x - 1\right) \left(x^{2} - 6 x - 2\right)}{2 x - 5}\right)$$
=
$$\lim_{x \to 1^+}\left(- \frac{\left(x - 1\right) \left(- x^{2} + 6 x + 2\right)}{2 x - 5}\right) = $$
$$- \frac{\left(-1 + 1\right) \left(- 1^{2} + 2 + 6\right)}{-5 + 2} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to 1^+}\left(\frac{4 x + \left(- 7 x^{2} + \left(x^{3} + 2\right)\right)}{2 x - 5}\right) = 0$$