Let $f(x)=\left\{\begin{array}{cc}(x-1) \sin \left(\frac{1}{x-1}\right), & \text { if } x \neq 1 \\ 0, & \text { if } x=1\end{array}\right.$ Then, which one of the following is true?
Answer & explanation
Correct answer: option 1
We observe that
$\lim\limits_{x \rightarrow 1} \frac{f(x)-f(1)}{x-1}$
$=\lim\limits_{x \rightarrow 1} \frac{(x-1) \sin \left(\frac{1}{x-1}\right)}{x-1}=\lim\limits_{x \rightarrow 1} \sin \left(\frac{1}{x-1}\right)$
= An oscillating number between -1 and 1.
∴ $\lim\limits_{x \rightarrow 1} \frac{f(x)-f(1)}{x-1}$ does not exist.
⇒ f(x) is not differentiable at x = 1.
and, $\lim\limits_{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}$
$=\lim\limits_{x \rightarrow 0} \frac{(x-1) \sin \left(\frac{1}{x-1}\right)-\sin 1}{x}$
$= \lim\limits_{x \rightarrow 0} \frac{x \sin \left(\frac{1}{x-1}\right)}{x}-\lim\limits_{x \rightarrow 0} \frac{\sin \left(\frac{1}{x-1}\right)+\sin 1}{x}$
$= -\sin 1-\lim\limits_{x \rightarrow 0} \frac{2 \sin \frac{x}{2(x-1)} \cos \left\{\frac{2-x}{2(x-1)}\right\}}{\left\{\frac{x}{2(x-1)}\right\} 2(x-1)}$
= - sin 1 + cos 1
⇒ f(x) is differentiable at x = 0.