Let f(x) = begin{cases} (x - 1) sin frac{1}{x | vedclass

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(D) To check differentiability at $x = 0$:$f'(0) = \lim_{h 	o 0} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0} \frac{(h-1) \sin(\frac{1}{h-1}) - (-1) \sin

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$f$ is differentiable at $x = 0$ but not $x = 1$

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(D) To check differentiability at $x = 0$:$f'(0) = \lim_{h 	o 0} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0} \frac{(h-1) \sin(\frac{1}{h-1}) - (-1) \sin(-1)}{h}$$= \lim_{h 	o 0} \frac{(h-1) \sin(\frac{1}{h-1}) - \sin(1)}{h}$.As $h 	o 0$,the limit is $\frac{-\sin(-1) - \sin(1)}{0}$,which is not defined. Thus,$f$ is differentiable at $x = 0$.To check differentiability at $x = 1$:$f'(1) = \lim_{h 	o 0} \frac{f(1+h) - f(1)}{h} = \lim_{h 	o 0} \frac{h \sin(\frac{1}{h}) - 0}{h} = \lim_{h 	o 0} \sin(\frac{1}{h})$.Since $\lim_{h 	o 0} \sin(\frac{1}{h})$ oscillates between $-1$ and $1$,the limit does not exist. Thus,$f$ is not differentiable at $x = 1$.Therefore,$f$ is differentiable at $x = 0$ but not at $x = 1$.

Let $f(x) = \begin{cases} (x - 1) \sin \frac{1}{x - 1}, & x \neq 1 \\ 0, & x = 1 \end{cases}$. Then which one of the following is true?

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