Question

Discuss the continuity and differentiability of

$f\left(x\right)=\left\{\begin{array}{ll}\left(x-c\right)\cos \left(\frac{1}{x-c}\right),& x\ne c\\ 0,& x=c\end{array}\right.$

Answer 1

Given: $f\left(x\right)=\left\{\begin{array}{ll}\left(x-c\right)\cos \left(\frac{1}{x-c}\right),& x\ne c\\ 0,& x=c\end{array}\right.$

Continuity:

(LHL at x = c)

$$\begin{gathered} \underset{x\to c^{-}}{\lim }f\left(x\right) \\ =\underset{h\to 0}{\lim }f(c-h) \\ =\underset{h\to 0}{\lim }(c-h-c)\cos \left(\frac{1}{c-h-c}\right) \\ =\underset{h\to 0}{\lim }-h\cos \left(\frac{1}{h}\right) \\ \mathrm{Since},\mathrm{co}s\left(\frac{1}{h}\right)\mathrm{is}\mathrm{a}\mathrm{bounded}\mathrm{function}\mathrm{and}0\times \mathrm{bounded}\mathrm{function}\mathrm{is}0 \\ =0 \end{gathered}$$

(RHL at x = c)

$$\begin{gathered} \underset{x\to c^{+}}{\lim }f\left(x\right) \\ =\underset{h\to 0}{\lim }f(c+h) \\ =\underset{h\to 0}{\lim }(c+h-c)\cos \left(\frac{1}{c+h-c}\right) \\ =\underset{h\to 0}{\lim }h\cos \left(\frac{1}{h}\right) \\ \mathrm{Since},\cos \left(\frac{1}{h}\right)\mathrm{is}\mathrm{a}\mathrm{bounded}\mathrm{function}\mathrm{and}0\times \mathrm{bounded}\mathrm{function}\mathrm{is}0 \\ =0 \end{gathered}$$

and
Differentiability at x = c

(LHD at x = c)
$$\begin{gathered} \underset{x\to c^{-}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c} \\ =\underset{h\to 0}{\lim }\frac{f(c-h)-f\left(c\right)}{c-h-c} \\ =\underset{h\to 0}{\lim }\frac{-h\cos \left({\displaystyle \frac{1}{-h}}\right)-0}{-h}\left[\because 0.\cos \left(\frac{1}{c-c}\right)=0,\mathrm{as}\cos \mathrm{function}\mathrm{is}\mathrm{bounded}\mathrm{function}.\right] \\ =\underset{h\to 0}{\lim }\cos \left(\frac{1}{h}\right) \\ =\mathrm{A}\mathrm{number}\mathrm{which}\mathrm{oscillates}\mathrm{between}-1\mathrm{and}1 \\ \therefore \mathrm{LHD}\hspace{0.17em}(x=c)\mathrm{does}\mathrm{not}\mathrm{exist}. \\ \mathrm{Similarly},\mathrm{we}\mathrm{can}\mathrm{show}\mathrm{that}\mathrm{RHD}(x=c)\mathrm{does}\mathrm{not}\mathrm{exist}. \\ \mathrm{Hence},\mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{not}\mathrm{differentiable}\mathrm{at}x=c \end{gathered}$$