Question

Consider the function $f\left(x\right)=\{\begin{array}{c}{x}^2\sin \frac{1}{x};x\ne 0\\ 0;otherwise\end{array}$
then,

• A. $f$ is derivable at $x=0$
• B. $f$ is not derivable at $x=0$
• C. $f$ is derivable at $x=0$ and $f^{\prime }\left(0\right)=0$
• D. $f$ is derivable at $x=0$ and $f^{\prime }\left(0\right)\ne 0$

Answer 1

The correct option is A $f$ is derivable at $x=0$

To check derivability at $x=0$

Concept : A function is derivable at $x=a$ if

$LHDat(x=a)=RHDat(x=a)$

$RHDatx=$

$=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}$

$=\underset{h\to 0}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}$

$=\underset{h\to 0}{lim}\frac{h^2\sin \left(\frac{1}{h}\right)-0}{h}$

$=\underset{h\to 0}{lim}h\sin \left(\frac{1}{h}\right)=0\times (valuebetween-1\&1)$

$=0$

Now

LHD at $x=0$.

$=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{-h}$

$=\underset{h\to 0}{lim}\frac{f(-h)-f\left(0\right)}{-h}$

$\underset{h\to 0}{lim}\frac{(-h{)}^2\sin (-\frac{1}{h})-0}{-h}$

$\underset{h\to 0}{lim}-h\sin (-\frac{1}{h})=\underset{h\to 0}{lim}h\sin \frac{1}{h}$

$=0\times [-1,1]$

$=0$

Here $\because LHD=RHD=0atx=0$

Hence f is derivable at $x=0$

Important Concept : Left Hand derivative $\left(LHD\right)=\underset{h\to 0}{lim}\frac{f(a-h)-f\left(a\right)}{-h}$

$atx=a$

$RHDat(x=a)=\underset{h\to 0}{lim}\frac{f(a+h)-f\left(a\right)}{h}$