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