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