Discuss the continuity and differentiability of

Given: $f(x)= \begin{cases}(x-c) \cos \left(\frac{1}{x-c}\right), & x \neq c \\ 0 & , x=c\end{cases}$

Continuity:

$(\mathrm{LHL}$ at $x=\mathrm{c})$

$\lim _{x \rightarrow c^{-}} f(x)$

$=\lim _{h \rightarrow 0} f(c-h)$

$=\lim _{h \rightarrow 0}(c-h-c) \cos \left(\frac{1}{c-h-c}\right)$

$=\lim _{h \rightarrow 0}-h \cos \left(\frac{1}{h}\right)$

Since, $\cos \left(\frac{1}{h}\right)$ is a bounded function and $0 \times$ bounded function is 0

$=0$

$\lim _{x \rightarrow c^{+}} f(x)$

$=\lim _{h \rightarrow 0} f(c+h)$

$=\lim _{h \rightarrow 0}(c+h-c) \cos \left(\frac{1}{c+h-c}\right)$

$=\lim _{h \rightarrow 0} h \cos \left(\frac{1}{h}\right)$

Since, $\cos \left(\frac{1}{h}\right)$ is a bounded function and $0 \times$ bounded function is 0

$=0$

and 
Differentiability at x = c

$(\mathrm{LHD}$ at $x=c)$

$\lim _{x \rightarrow c^{-}} \frac{f(x)-f(c)}{x-c}$

$=\lim _{h \rightarrow 0} \frac{f(c-h)-f(c)}{c-h-c}$

$=\lim _{h \rightarrow 0} \frac{-h \cos \left(\frac{1}{-h}\right)-0}{-h}$    $\left[\because 0 . \cos \left(\frac{1}{c-c}\right)=0\right.$, as $\cos$ function is bounded function. $]$

$=\lim _{h \rightarrow 0} \cos \left(\frac{1}{h}\right)$

$\therefore \operatorname{LHD}(x=c)$ does not exist.

Similarly, we can show that $\mathrm{RHD}(x=c)$ does not exist.

Hence, $\mathrm{f}(\mathrm{x})$ is not differentiable at $x=c$