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