Question

$\underset{x\to 1}{lim}\frac{x\sin \left\{x\right\}}{x-1},$ where {x} denotes the fractional part of x, is equal to

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

Answer 1

The correct option is D does not exist


$\underset{x\to 1-0}{lim}\left\{x\right\}=\underset{x\to 1-0}{lim}(x-[x\left]\right)$ = 1-0=1
$\underset{x\to 1+0}{lim}\left\{x\right\}=\underset{x\to 1+0}{lim}(x-[x\left]\right)$ = 1-1=0
$\therefore \underset{x\to 1-0}{lim}\frac{x\sin \left\{x\right\}}{x-1}=\underset{x\to 1-0}{lim}\sin \left\{x\right\}=-\infty$
$\underset{x\to 1+0}{lim}\frac{x\sin \left\{x\right\}}{\left\{x\right\}}\frac{x-1}{x-1}=1\times 1\times 1=1$
Since, L.H. limit $\ne$ R.H. limit Limit does not exist.