Question

If $f\left(x\right)=\{\begin{array}{cc}x\sin \left(\frac{1}{x}\right),& x\ne 0\\ 0,& x=0\end{array}$, then at $x=0$ the function $f\left(x\right)$ is

• A. Continuous
• B. Differentiable
• C. Continuous but not differentiable
• D. None of the above

Answer 1

The correct option is C Continuous but not differentiable

Given, $f\left(x\right)=\{\begin{array}{cc}x\sin \left(\frac{1}{x}\right),& x\ne 0\\ 0,& x=0\end{array}$
For continuity at $x=0$,
$LHL=f(0-0)=\underset{h\to 0}{lim}f(0-h)$
$=\underset{h\to 0}{lim}(-h)\sin (-\frac{1}{h})$
$=0\times \left(\text{finite quantity}\right)$
$=0$
$RHL=f(0+0)=\underset{h\to 0}{lim}f(0+h)$$
$=\underset{h\to 0}{lim}\left(h\right)\sin \left(\frac{1}{h}\right)$
$=0\times \left(\text{finite quantity}\right)$
$=0$
and $f\left(0\right)=0$
$\therefore f\left(0\right)=LHL=RHL$
$\therefore f\left(x\right)$ is continuous at $x=0$.
For differentiability at $x=0$,
$R{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}$
$=\underset{h\to 0}{lim}\frac{h\sin \left(\frac{1}{h}\right)-0}{h}$
$=\underset{h\to 0}{lim}\sin \left(\frac{1}{h}\right)$
$=$ (a finite quantity persist between $-1$ to $+1$)
$=$ does not exist
$L{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{-h}$
$=\underset{h\to 0}{lim}\frac{(-h)\sin (-\frac{1}{h})-0}{-h}$
$=\underset{h\to 0}{lim}\{-\sin \left(\frac{1}{h}\right)\}$
$=$ (finite quantity persist between $-1$ to $+1$)
$=$ does not exist
$\because R{f}^{\prime }\left(0\right)\ne L{f}^{\prime }\left(0\right)$
$\therefore f\left(x\right)$ is not differentiable at $x=0$.