Continuity and Differentiability — Class 12 Maths Solution

We have $$f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{|x - a|\sin \frac{1}{{x - a}},}&{{\rm{ if }}x \ne 0}\\{0,}&{{\rm{ if }}x = a}\end{array}} \right.$$ at $x = a$
At $x = a,$ ${\rm{LHL}} = \mathop {\lim }\limits_{x \to {a^ - }} |x - a|\sin \frac{1}{{x - a}}$
$= \mathop {\lim }\limits_{h \to 0} |a - h - a|\sin \left( {\frac{1}{{a - h - a}}} \right)$
$= \mathop {\lim }\limits_{h \to 0} - h\sin \left( {\frac{1}{h}} \right)$
$= 0 \times [$ an oscillating number between -1 and 1$] = 0$
${\rm{RHL}} = \mathop {\lim }\limits_{x \to {a^ + }} |x - a|\sin \left( {\frac{1}{{x - a}}} \right)$
$= \mathop {\lim }\limits_{h \to 0} |a + h - a|\sin \left( {\frac{1}{{a + h - a}}} \right) = \mathop {\lim }\limits_{h \to 0} h\sin \frac{1}{h}$
$= 0 \times [$ an oscillating number between -1 and 1$] = 0$
and $f(a) = 0$

$\Rightarrow {\rm{LHL}} = {\rm{RHL}} = f({\rm{a}})$

Therefore we can say that $f(x)$ is continuous at $x = a$.