$f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{|x|\cos \frac{1}{x},}&{{\rm{ if }}x \ne 0}\\{0,}&{{\rm{ if }}x = 0}\end{array}} \right.$at $x = 0$ }

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

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