Question

$f\left(x\right)=x^{m}Sin\left(\frac{1}{4}\right)$ if x $\ne$ 0
0 if x=0 is continuous but not differentiable at x=0 then find the value of 'm'.

Answer 1

$f\left(x\right)=\{\begin{array}{c}{x}^m\sin \left(\frac{1}{4}\right);x\ne 0\\ 0;x=0\end{array}$

$\underset{x\to 0}{lim}f\left(x\right)=\sin \left(\frac{1}{4}\right)\underset{x\to 0}{lim}{x}^m=\sin \left(\frac{1}{4}\right)\times 0^m$

$f\left(0\right)$should be equal to ${lim}_{x\to 0^{-}}f\left(x\right)$

$\Rightarrow m>0\to 1$

$f^{\prime }\left(x\right)=\{\begin{array}{c}m{x}^{m-1}\sin \left(\frac{1}{4}\right);x\ne 0\\ 0;x=0\end{array}$

$\underset{x\to 0}{lim}{f}^{\prime }\left(x\right)=m\sin \left(\frac{1}{4}\right)\underset{x\to 0}{lim}{x}^{m-1}$

$=m\sin \left(\frac{1}{4}\right)\times 0^{m-1}$

But it is not differentiable at $x=0$

$\Rightarrow 0^{m-1}\ne 0\Rightarrow m-1\le 0$

$\Rightarrow m\le 1\to 2$

From $1$ and $2$

$m=1$