Tomamos como el límite
$$\lim_{x \to 2^+}\left(\frac{5 x^{2} + \left(1 - 6 x\right)}{1 - x^{2}}\right)$$
cambiamos
$$\lim_{x \to 2^+}\left(\frac{5 x^{2} + \left(1 - 6 x\right)}{1 - x^{2}}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(x - 1\right) \left(5 x - 1\right)}{\left(-1\right) \left(x - 1\right) \left(x + 1\right)}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{1 - 5 x}{x + 1}\right) = $$
$$\frac{1 - 10}{1 + 2} = $$
= -3
Entonces la respuesta definitiva es:
$$\lim_{x \to 2^+}\left(\frac{5 x^{2} + \left(1 - 6 x\right)}{1 - x^{2}}\right) = -3$$