Question

Let the function f, g and h be defined as follows :
$f\left(x\right)=x\sin \left(\frac{1}{x}\right)$ For $-1\le x\le 1$ and $x\ne 0$ $0$ For $x=0$
$g\left(x\right)=x^2\sin \left(\frac{1}{x}\right)$ For $-1\le x\le 1$ and $x\ne 0$ $0$ For $x=0$
$h\left(x\right)={\left|x\right|}^3$ For $-1\le x\le 1$
Which of these functions are differentiable at $x=0?$

• A. $f$ and $g$ only
• B. $f$ and $h$ only
• C. $g$ and $h$ only
• D. none

Answer 1

Answer: (C)

$g$ and $h$ only

(1) $f\left(x\right)=x\sin \left(\frac{1}{x}\right)$ For $-1\le x\le 1$ and $x\ne 0$ $0$ For $x=0$

$f\left(x\right)$ is not differentiable at $x=0$

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

which does not exist.

(2) $g\left(x\right)=x^2\sin \left(\frac{1}{x}\right)$ For $-1\le x\le 1$ and $x\ne 0$ $0$ For $x=0$

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

Similarly $L{f}^{\prime }\left(0\right)=0$

Hence, $g\left(x\right)$ is differentiable at $x=0$.

(3) $h\left(x\right)={\left|x\right|}^3$ For $-1\le x\le 1$

$RHD=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}=\underset{h\to 0}{lim}\frac{|h{|}^3-0}{h}=\underset{h\to 0}{lim}{h}^2=0$

$LHD=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{-h}=\underset{h\to 0}{lim}\frac{|-h{|}^3-0}{-h}=\underset{h\to 0}{lim}-h^2=0$

Since $f^{\prime }\left(0\right)=RHD=LHD=0$, $h\left(x\right)$ is differentiable at $x=0$.

Hence, only $g$ and $h$ are differentiable.

Hence, option C is correct.