# Show that the function

Question:

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

(i) differentiable at $x=0$, if $m>1$

(ii) continuous but not differentiable at $x=0$, if $0 (iii) neither continuous nor differentiable, if$m \leq 0$Solution: Given:$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}\$

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