Question

For each $t\in R$, let $\left[t\right]$ be the greatest integer less than or equal to t. Then, $\underset{x\to 1+}{lim}\frac{(1-|x|+\sin |1-x|)\sin \left(\frac{\pi }{2}\right[1-x\left]\right)}{|1-x|[1-x]}$.

• A. Equals $-1$
• B. Equals $1$
• C. Does not exist
• D. Equals $0$

Answer 1

The correct option is D Equals $0$

$\underset{x\to 1^{+}}{lim}\frac{(1-|x|+\sin |1-x|)\sin \left(\frac{\pi }{2}\right[1-x\left]\right)}{|1-x|[1-x]}$
$=\underset{x\to 1^{+}}{lim}\frac{(1-x)+\sin (x-1)}{(x-1)(-1)}\sin \left(\frac{\pi }{2}\right(-1\left)\right)$
$=\underset{x\to 1^{+}}{lim}(1-\frac{\sin (x-1)}{(x-1)})(-1)=(1-1)(-1)=0$.