Tomamos como el límite
$$\lim_{x \to -1^+}\left(\frac{3 x + \left(x^{2} - 28\right)}{x^{4} - 256}\right)$$
cambiamos
$$\lim_{x \to -1^+}\left(\frac{3 x + \left(x^{2} - 28\right)}{x^{4} - 256}\right)$$
=
$$\lim_{x \to -1^+}\left(\frac{\left(x - 4\right) \left(x + 7\right)}{\left(x - 4\right) \left(x + 4\right) \left(x^{2} + 16\right)}\right)$$
=
$$\lim_{x \to -1^+}\left(\frac{x + 7}{\left(x + 4\right) \left(x^{2} + 16\right)}\right) = $$
$$\frac{-1 + 7}{\left(-1 + 4\right) \left(\left(-1\right)^{2} + 16\right)} = $$
= 2/17
Entonces la respuesta definitiva es:
$$\lim_{x \to -1^+}\left(\frac{3 x + \left(x^{2} - 28\right)}{x^{4} - 256}\right) = \frac{2}{17}$$