Question

Let the function $f\left(x\right)=\frac{\left\{x\right\}}{\tan \left\{x\right\}}.$ Then
( where $\left\{x\right\}$ denotes fractional part of $x$ )

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

Answer 1

Answer: (A), (C)

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

Given : $f\left(x\right)=\frac{\left\{x\right\}}{\tan \left\{x\right\}}$

$L.H.L.=\underset{x\to 0^{-}}{lim}\frac{\left\{x\right\}}{\tan \left\{x\right\}}$
$=\underset{h\to 0}{lim}\frac{\{0-h\}}{\tan \{0-h\}}=\underset{h\to 0}{lim}\frac{\{-h\}}{\tan \{-h\}}$
$=\frac{1}{\tan 1}$
$R.H.L.=\underset{x\to 0^{+}}{lim}\frac{\left\{x\right\}}{\tan \left\{x\right\}}$
$=\underset{h\to 0}{lim}\frac{\{0+h\}}{\tan \{0+h\}}=\underset{h\to 0}{lim}\frac{h}{\tan h}$
$=1$
Clearly, $L.H.L.\ne R.H.L.$
$\therefore$ $f\left(x\right)$ is discontinuous at $x=0$