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

Answer 1

The correct option is A equals $0$

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