If \(f(x)=\left\{\begin{array}{c}x \sin \left(\frac{1}{x}\right), x \neq 0 \text { , then at } x=0 \text { the } \ 0, x=0\end{array}\right.\)
function \(f(x)\) is

Given, \(f(x)=\left\{\begin{array}{c}x \sin \left(\frac{1}{x}\right), x \neq 0 \\ 0, \quad x=0\end{array}\right.\)
For continuity at \(x=0\),
\(\mathrm{LHL}=f(0-0)=\lim _{h \rightarrow 0} f(0-h)\)
\(=\lim _{h \rightarrow 0}(-h) \sin \left(-\frac{1}{h}\right) \)
\( =0 \times(\text { finite quantity }) \)
\( =0 \)
\( \text { RHL } =f(0+0)=\lim _{h \rightarrow 0} f(0+h) \)
\( =\lim _{h \rightarrow 0}(h) \sin \left(\frac{1}{h}\right)\)
\(=0 \times(\) finite quantity \()\)
\(=0\)
and \(f(0)=0\)
\(\therefore \quad f(0)=\mathrm{LHL}=\mathrm{RHL}\)
\(\therefore f(x)\) is continuous at \(x=0 .\)
For differentiability at \(x=0\),
\(R f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}\)
\(=\lim _{h \rightarrow 0} \frac{h \sin \left(\frac{1}{h}\right)-0}{h} \)
\( =\lim _{h \rightarrow 0} \sin \left(\frac{1}{h}\right) \)
\( =(\text { a finite quantity persist between }-1\)
\( \text { to }+1) \)
\( =\text { does not exist } \)
\( L f^{\prime}(0) =\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h} \)
\( =\lim _{h \rightarrow 0} \frac{(-h) \sin \left(-\frac{1}{h}\right)-0}{-h} \)
\(= \lim _{h \rightarrow 0}\left\{-\sin \left(\frac{1}{h}\right)\right\} \)
\(=(\text { finite quantity persist between }\)
\( -1 \text { to }+1) \)
\( =\text { does not exist } \)
\( \because R f^{\prime}(0) \neq L^{\prime}(0)\)
\(\therefore f(x)\) is not differentiable at \(x=0\).