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Let $f(x)=\left\{\begin{array}{ll}(x-1) \sin \left(\frac{1}{x-1}\right), & \text { if } x \neq 1 \\ 0, & \text { if } x=1\end{array}\right.$. Then which one of the following is true?
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The correct answer is:
$f$ is neither differentiable at $x=0$ nor at $x=1$
$f$ is neither differentiable at $x=0$ nor at $x=1$
$$
f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}
$$
$$
\begin{aligned}
& \Rightarrow f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{(1+h-1) \sin \left(\frac{1}{1+h-1}\right)-0}{h}=\lim _{h \rightarrow 0} \frac{h}{h} \sin \left(\frac{1}{h}\right) \\
& \Rightarrow f^{\prime}(1)=\lim _{h \rightarrow 0} \sin \left(\frac{1}{h}\right)
\end{aligned}
$$
$\therefore \mathrm{f}$ is not differentiable at $x=1$.
Similarly, $f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}$ $\Rightarrow f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{(h-1) \sin \left(\frac{1}{h-1}\right)-\sin (1)}{h}$
$\Rightarrow \mathrm{f}$ is also not differentiable at $\mathrm{x}=0$.
f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}
$$
$$
\begin{aligned}
& \Rightarrow f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{(1+h-1) \sin \left(\frac{1}{1+h-1}\right)-0}{h}=\lim _{h \rightarrow 0} \frac{h}{h} \sin \left(\frac{1}{h}\right) \\
& \Rightarrow f^{\prime}(1)=\lim _{h \rightarrow 0} \sin \left(\frac{1}{h}\right)
\end{aligned}
$$
$\therefore \mathrm{f}$ is not differentiable at $x=1$.
Similarly, $f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}$ $\Rightarrow f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{(h-1) \sin \left(\frac{1}{h-1}\right)-\sin (1)}{h}$
$\Rightarrow \mathrm{f}$ is also not differentiable at $\mathrm{x}=0$.
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