Tomamos como el límite
$$\lim_{x \to 1^+}\left(\frac{- 2 x + \left(x^{2} - 8\right)}{2 x^{2} + \left(3 - 8 x\right)}\right)$$
cambiamos
$$\lim_{x \to 1^+}\left(\frac{- 2 x + \left(x^{2} - 8\right)}{2 x^{2} + \left(3 - 8 x\right)}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{\left(x - 4\right) \left(x + 2\right)}{2 x^{2} - 8 x + 3}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{\left(x - 4\right) \left(x + 2\right)}{2 x^{2} - 8 x + 3}\right) = $$
$$\frac{\left(-4 + 1\right) \left(1 + 2\right)}{-8 + 2 \cdot 1^{2} + 3} = $$
= 3
Entonces la respuesta definitiva es:
$$\lim_{x \to 1^+}\left(\frac{- 2 x + \left(x^{2} - 8\right)}{2 x^{2} + \left(3 - 8 x\right)}\right) = 3$$