Question

is continuous but not derivable at $x=0$

• A. $m\in [0,3]$
• B. $m\in (0,2)$
• C. $m\in (0,1]$
• D. $m=0,1$

Answer 1

Answer: (C)

$m\in (0,1]$

$f\left(0^{+}\right)=\frac{{lim}_{h\to 0}{(0+h)}^m\sin \left(\frac{1}{h}\right)-0}{h}$
$=\underset{h\to 0}{lim}{h}^{m-1}\sin \left(\frac{1}{h}\right)$
$\Rightarrow m-1>0$ for differentiability
$\Rightarrow m\in (0,1],f\left(x\right)$ is continuous but not derivable.