If f(x) = begin{cases} x sin left(frac{1}{x}r | vedclass

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(C) Given,$f(x) = \begin{cases} x \sin \left(\frac{1}{x}ight), & x 
eq 0 \ 0, & x = 0 \end{cases}$ For continuity at $x = 0$: $LHL = f(0-0) = \lim

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continuous but not differentiable

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(C) Given,$f(x) = \begin{cases} x \sin \left(\frac{1}{x}ight), & x 
eq 0 \ 0, & x = 0 \end{cases}$ For continuity at $x = 0$: $LHL = f(0-0) = \lim_{h 	o 0} f(0-h) = \lim_{h 	o 0} (-h) \sin \left(-\frac{1}{h}ight) = \lim_{h 	o 0} h \sin \left(\frac{1}{h}ight) = 0 	imes (	ext{finite quantity}) = 0$. $RHL = f(0+0) = \lim_{h 	o 0} f(0+h) = \lim_{h 	o 0} h \sin \left(\frac{1}{h}ight) = 0 	imes (	ext{finite quantity}) = 0$. Since $f(0) = LHL = RHL = 0$,$f(x)$ is continuous at $x = 0$. For differentiability at $x = 0$: $Rf'(0) = \lim_{h 	o 0} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0} \frac{h \sin(1/h) - 0}{h} = \lim_{h 	o 0} \sin(1/h)$. As $h 	o 0$,$\sin(1/h)$ oscillates between $-1$ and $1$,so the limit does not exist. Similarly,$Lf'(0) = \lim_{h 	o 0} \frac{f(0-h) - f(0)}{-h} = \lim_{h 	o 0} \frac{-h \sin(-1/h) - 0}{-h} = \lim_{h 	o 0} \sin(1/h)$,which also does not exist. Since the derivative does not exist,$f(x)$ is not differentiable at $x = 0$. Thus,$f(x)$ is continuous but not differentiable at $x = 0$.

If $f(x) = \begin{cases} x \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}$,then at $x = 0$ the function $f(x)$ is

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