Question

Let the function $f,g$ and $h$ be defined as follows-
$f\left(x\right)=\{\begin{array}{c}x\sin \left(\frac{1}{x}\right)-1\le x\le 1\text{and}x\ne 0\\ 0x=0\end{array}$, $g\left(x\right)=\{\begin{array}{c}{x}^2\sin \left(\frac{1}{x}\right)-1\le x\le 1\text{and}x\ne 0\\ 0x=0\end{array}h\left(x\right)={\left|x\right|}^3-1\le x\le 1$
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

$\left(i\right)f\left(x\right)=\{\begin{array}{c}x\sin \frac{1}{x};-1\le x\le 1\\ 0;x=0\end{array}$

LHD$={lim}_{h\to 0^{-}}\frac{h\sin \frac{1}{h}-0}{h-0}={lim}_{h\to 0}\frac{h\sin \frac{1}{h}}{h}={lim}_{h\to 0}\sin \frac{1}{h}=$ does not exists.

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

$\left(ii\right)g\left(x\right)=\{\begin{array}{c}{x}^2\sin \frac{1}{x};-1\le x\le 1\\ 0;x=0\end{array}$

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

RHD$={lim}_{h\to 0^{+}}\frac{h^2\sin \frac{1}{h}-0}{h-0}={lim}_{h\to 0}\frac{h^2\sin \frac{1}{h}}{h}={lim}_{h\to 0}h\sin \frac{1}{h}=0$

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

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

LHD$={lim}_{h\to 0^{-}}\frac{{|-h|}^3-0}{h-0}={lim}_{h\to 0}{h}^2=0$

Hence $h\left(x\right)$ is differentiable at $x=0$