Tomamos como el límite
$$\lim_{x \to 1^+}\left(\frac{- 6 x + \left(x^{2} + 8\right)}{- 5 x + \left(x^{2} + 4\right)}\right)$$
cambiamos
$$\lim_{x \to 1^+}\left(\frac{- 6 x + \left(x^{2} + 8\right)}{- 5 x + \left(x^{2} + 4\right)}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{\left(x - 4\right) \left(x - 2\right)}{\left(x - 4\right) \left(x - 1\right)}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{x - 2}{x - 1}\right) = $$
False
= -oo
Entonces la respuesta definitiva es:
$$\lim_{x \to 1^+}\left(\frac{- 6 x + \left(x^{2} + 8\right)}{- 5 x + \left(x^{2} + 4\right)}\right) = -\infty$$