Question

If $\left[x\right]$ is the greatest integer less than or equal to $x$, for all $x\in R$, then evaluate $\underset{x\to 1^{+}}{lim}\frac{1-|x|+\sin |1-x|}{|1-x|[1-x]}$

• A. $3$
• B. $0$
• C. $6$
• D. $-5$

Answer 1

The correct option is B $0$

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