Question

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

• A. $\sin 1$
• B. $1$
• C. $-1$
• D. limit does not exist.

Answer 1

The correct option is D limit does not exist.

Given : $f\left(x\right)=\{\begin{array}{cc}\frac{\sin \left[x\right]}{\left[x\right]},& x\ne 0\\ 0,& x=0\end{array}$

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

$R.H.L=\underset{x\to 0^{+}}{lim}f\left(x\right)$
$=\underset{h\to 0}{lim}f(0+h)$
$=\underset{h\to 0}{lim}\frac{\sin \left[h\right]}{\left[h\right]}$
which is not defined because $h\to 0\Rightarrow \left[h\right]=0$

$\therefore \underset{x\to 0}{lim}f\left(x\right)$ does not exist.