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