Tomamos como el límite
$$\lim_{x \to 2^+}\left(\frac{4 - x^{2}}{\left(8 - 8 x\right) - 4 \sqrt{5}}\right)$$
cambiamos
$$\lim_{x \to 2^+}\left(\frac{4 - x^{2}}{\left(8 - 8 x\right) - 4 \sqrt{5}}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(-1\right) \left(x - 2\right) \left(x + 2\right)}{- 8 x - 4 \sqrt{5} + 8}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{x^{2} - 4}{4 \left(2 x - 2 + \sqrt{5}\right)}\right) = $$
$$\frac{-4 + 2^{2}}{4 \left(-2 + \sqrt{5} + 2 \cdot 2\right)} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to 2^+}\left(\frac{4 - x^{2}}{\left(8 - 8 x\right) - 4 \sqrt{5}}\right) = 0$$