Let $$f(x) = \left\{ {\matrix{ {{x^2}\sin \left( {{1 \over x}} \right)} & {,\,x \ne 0} \cr 0 & {,\,x = 0} \cr } } \right.$$

Then at $x=0$

<p>Given,</p> <p>$$f(x) = \left\{ {\matrix{ {{x^2}\sin \left( {{1 \over x}} \right),} & {x \ne 0} \cr {0,} & {x = 0} \cr } } \right.$$</p> <p>$\therefore$ $f'(x) = 2x\sin \left( {{1 \over x}} \right) - \cos \left( {{1 \over x}} \right)$</p> <p>Now,</p> <p>$$\mathop {\lim }\limits_{x \to 0} f'(x) = \mathop {\lim }\limits_{x \to 0} \left[ {2x\sin \left( {{1 \over x}} \right) - \cos \left( {{1 \over x}} \right)} \right]$$</p> <p>$= 0 - \mathop {\lim }\limits_{x \to 0} \cos \left( {{1 \over x}} \right)$</p> <p>= Does not exist</p> <p>$\therefore$ $f'(x)$ is discontinuous function at $x = 0$.</p> <p>L.H.D. $= \mathop {\lim }\limits_{h \to {0^ + }} {{f(0 - h) - f(0)} \over { - h}}$</p> <p>$= \mathop {\lim }\limits_{h \to {0^ + }} {{f( - h)} \over { - h}}$</p> <p>$$ = \mathop {\lim }\limits_{h \to {0^ + }} {{ - {h^2}\sin \left( {{1 \over h}} \right)} \over { - h}} = 0$$</p> <p>R.H.D. $= \mathop {\lim }\limits_{h \to {0^ + }} {{f(0 + h) - f(0)} \over h}$</p> <p>$= \mathop {\lim }\limits_{h \to {0^ + }} {{f(h)} \over h}$</p> <p>$$ = \mathop {\lim }\limits_{h \to {0^ + }} {{{h^2}\sin \left( {{1 \over h}} \right)} \over h} = 0$$</p> <p>$\therefore$ L.H.D. = R.H.D.</p> <p>$\Rightarrow f(x)$ is differentiable at $x = 0$. So, f(x) is continuous.</p>