Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{\sin \left[x\right]}{\left[x\right]},& \left[x\right]\ne 0\\ 0,& \left[x\right]=0\end{array}$ where $\left[x\right]$ denotes the greatest integer less than or equal to $x$, then $\underset{x\to 0}{lim}f\left(x\right)$ equals.?

• A. $1$
• B. $0$
• C. $-1$
• D. None of these

Answer 1

The correct option is D None of these

As, $f\left(x\right)=\{\begin{array}{cc}\frac{\sin \left[x\right]}{\left[x\right]},& \left[x\right]\ne 0\\ 0,& \left[x\right]=0\end{array}$
$\Rightarrow f\left(x\right)=\{\begin{array}{cc}\frac{\sin \left[x\right]}{\left[x\right]},& xϵR-[0,1)\\ 0,& 0\le x<1\end{array}$
$\therefore$ RHL at $x=0$
$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{\sin [0+h]}{[0+h]}=0$
$\text{LHL}$ at $x=0$
$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{\sin [0-h]}{[0-h]}$
$=\underset{h\to 0}{lim}\frac{\sin (-1)}{-1}=\sin 1$
Since, $\text{RHL}\ne \text{LHL}$
Therefore, limit does not exist.