Question:
For each $t \in R$, let [t] be the greatest integer less than or equal to $\mathrm{t}$. Then,
$\lim _{x \rightarrow 1+} \frac{(1-|x|+\sin |1-x|) \sin \left(\frac{\pi}{2}[1-x]\right)}{|1-x|[1-x]}$
Correct Option: , 4
Solution:
$\lim _{x \rightarrow 1^{+}} \frac{(1-|x|+\sin |1-x|) \sin \left(\frac{\pi}{2}[1-x]\right)}{|1-x|[1-x]}$
$=\lim _{x \rightarrow 1^{+}} \frac{(1-x)+\sin (x-1)}{(x-1)(-1)} \sin \left(\frac{\pi}{2}(-1)\right)$
$=\lim _{x \rightarrow 1^{+}}\left(1-\frac{\sin (x-1)}{(x-1)}\right)(-1)=(1-1)(-1)=0$