$$\dfrac{x-1}{x^2-1} = \dfrac{\cancel{x-1}}{\cancel{(x-1)}\cdot (x+1)} = \dfrac{1}{x+1}$$ $$\displaystyle{\lim_{x \rightarrow 1} \left(\dfrac{x-1}{x^2-1}\right)} = \displaystyle{\lim_{x \rightarrow 1} \left(\dfrac{1}{x+1}\right)} = \dfrac{1}{2} = 0,5$$ $$\pu{Alternativa } \mathbb{(D)}$$