Show that the function

Given:

$f(x)= \begin{cases}x^{m} & \sin \left(\frac{1}{x}\right) \\ 0 & \mathrm{x} \neq 0, \mathrm{x}=0\end{cases}$

(i) Let $\mathrm{m}=2$, then the function becomes $f(x)= \begin{cases}x^{2} & \sin \left(\frac{1}{x}\right) \\ 0 & , \mathrm{x} \neq 0, \mathrm{x}=0\end{cases}$

Differentiability at x=0:

$\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0} \frac{f(x)}{x}=\lim _{x \rightarrow 0} x \sin \left(\frac{1}{x}\right)=0$

$\left[\because \lim _{x \rightarrow 0} x \sin \left(\frac{1}{x}\right)=0\right.$, as $\left|x \sin \frac{1}{x}-0\right|=\left|x \sin \frac{1}{x}\right|=|x|\left|\sin \frac{1}{x}\right| \leq|x| \quad\left(\because|\sin \theta| \leq 1\right.$ for all $\theta$ ) and hence $\mid x$ sin $\frac{1}{x} \mid<0$ when

$|x-0|<\varepsilon|x-0|<\varepsilon]$

So, $f^{\prime}(0)=0$, which means $f$ is differentiable at $x=0$.

Hence the given function is differentiable at $x=0$.

(ii) Let $m=\frac{1}{2}, 0

$f(x)=\left\{\begin{array}{l}x^{\frac{1}{2}} \\ 0\end{array} \sin \left(\frac{1}{x}\right)\right.$ $x \neq 0, x=0$

Continuity at x=0:

(LHL at $\mathrm{x}=0$ ) $=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0}(-h)^{\frac{1}{2}} \sin \left(\frac{1}{0-h}\right)=\lim _{h \rightarrow 0} h^{\frac{1}{2}} \sin \left(\frac{1}{h}\right)=\lim _{h \rightarrow 0} h^{\frac{3}{2}}=0$.

(RHL at $\mathrm{x}=0$ ) $=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} h^{\frac{1}{2}} \sin \left(\frac{1}{h}\right)=\lim _{h \rightarrow 0} h^{\frac{3}{2}}=0$.

and $f(0)=0$

LHL at $\mathrm{x}=0=\mathrm{RHL}$ at $\mathrm{x}=0=\lim _{x \rightarrow 0} f(x)$,

Hence continuous.

Now Differentiabilty at $x=0$ when $0

$($ LHD at $x=0)=\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{0-h-0}=\lim _{h \rightarrow 0} \frac{(-h)^{\frac{1}{2}} \sin \left(\frac{1}{-h}\right)}{-h}$