Question

If $\left\{x\right\}$ denotes fractional part of $x$, then $\underset{x\to 1}{lim}\frac{x\sin \left\{x\right\}}{x-1}=$

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

Answer 1

The correct option is D does not exist

As $x\to 1^{+};\left\{x\right\}=1$ and as $x\to 1^{-};\left\{x\right\}=0$

$\therefore L.H.L=\underset{x\to 1^{-}}{lim}\frac{x\sin \left\{x\right\}}{x-1}=\underset{x\to 1^{-}}{lim}\frac{x\sin \left(0\right)}{x-1}=\underset{x\to 1^{-}}{lim}0=0$

And $R.H.L=\underset{x\to 1^{+}}{lim}\frac{x\sin \left\{x\right\}}{x-1}=\underset{x\to 1^{+}}{lim}\frac{x\sin \left(1\right)}{x-1}$

Applying L'Hospital rule

$R.H.L=\underset{x\to 1^{+}}{lim}\frac{\sin \left(1\right)}{-1}=-\sin \left(1\right)$

$\because L.H.L\ne R.H.L$

$\therefore$ limit at $x=1$ does no exists.