Tomamos como el límite
$$\lim_{x \to 1^+}\left(\frac{2 x + \left(x^{3} - 1\right)}{- 7 x^{3} - 4 x^{2}}\right)$$
cambiamos
$$\lim_{x \to 1^+}\left(\frac{2 x + \left(x^{3} - 1\right)}{- 7 x^{3} - 4 x^{2}}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{x^{3} + 2 x - 1}{\left(-1\right) x^{2} \left(7 x + 4\right)}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{- x^{3} - 2 x + 1}{x^{2} \left(7 x + 4\right)}\right) = $$
$$\frac{-2 - 1^{3} + 1}{4 + 7} = $$
= -2/11
Entonces la respuesta definitiva es:
$$\lim_{x \to 1^+}\left(\frac{2 x + \left(x^{3} - 1\right)}{- 7 x^{3} - 4 x^{2}}\right) = - \frac{2}{11}$$