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

• A. $f\left(x\right)$ is continuous at $x=0$
• B. $f\left(x\right)$ is discontinuous at $x=0$
• C. $\underset{x\to 0^{-}}{lim}f\left(x\right)=\sin 1$
• D. $\underset{x\to 0^{+}}{lim}f\left(x\right)=\sin 1$

Answer 1

Answer: (B), (C)

$\underset{x\to 0^{-}}{lim}f\left(x\right)=\sin 1$

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.
$\therefore$ $f\left(x\right)$ is discontinuous at $x=0$