Continuity and Differentiability — Class 12 Maths Solution

We have, $$f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{{x^2}\sin \frac{1}{x},}&{{\rm{ if }}x \ne 0}\\{0,}&{{\rm{ if }}x = 0}\end{array}} \right.$$at $x = 0$

For differentiability at $x = 0$

$L{f^\prime }(0) = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{f(x) - f(0)}}{{x - 0}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{{x^2}\sin \frac{1}{x} - 0}}{{x - 0}}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{{{(0 - h)}^2}\sin \left( {\frac{1}{{0 - h}}} \right)}}{{0 - h}} = \mathop {\lim }\limits_{h \to 0} \frac{{{h^2}\sin \left( {\frac{{ - 1}}{h}} \right)}}{{ - h}}$
$= \mathop {\lim }\limits_{h \to 0} + h\sin \left( {\frac{1}{h}} \right)$

$= 0 \times [$ some constant which is oscillating between -1 and 1 $] = 0$

$R{f^\prime }(0) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{f(x) - f(0)}}{{x - 0}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{{x^2}\sin \frac{1}{x} - 0}}{{x - 0}}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{{{(0 + h)}^2}\sin \left( {\frac{1}{{0 + h}}} \right)}}{{0 + h}} = \mathop {\lim }\limits_{h \to 0} \frac{{{h^2}\sin (1/h)}}{h}$

$= \mathop {\lim }\limits_{h \to 0} h\sin (1/h)$

$= 0 \times [$ some constant which is oscillating between -1 and 1$] = 0$

$\Rightarrow L{f^\prime }(0) = R{f^\prime }(0)$

Therefore we can say that $f(x)$ is differentiable at $x = 0$.