For each $t\in\mathbb{R}$, let $[t]$ be the greatest integer less than or equal to $t$. Then $\displaystyle \lim_{x\to 1^{+}}\frac{\big(1-|x|+|\sin|1-x||\big)\,\sin\!\left(\tfrac{\pi}{2}[\,1-x\,]\right)}{|1-x|\,[\,1-x\,]}$ is: