Question

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

• A. equals -1
• B. equals 1
• C. does not exist
• D. equals 0

Answer 1

Answer: (D)

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