Discuss the continuity and differentiability of
$f(x)= \begin{cases}(x-c) \cos \left(\frac{1}{x-c}\right), & x \neq c \\ 0 & , x=c\end{cases}$
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 $x=c$
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