Question

Let $f(x)=\{\begin{array}{ll}(x-1)\sin \frac{1}{x-1}& \text{if }x\ne 1\\ 0& \text{if }x=1\end{array}$
Then which one of the following is true?

• A. f is neither differentiable at x = 0 nor at x =1
• B. f is differentiable at x = 0 and at x = 1
• C. f is differentiable at x = 0 but not at x = 1
• D. f is differentiable at x = 1 but not at x = 0

Answer 1

Answer: (C)

We have $f(x)=\{\begin{array}{ll}(x-1)\sin \left(\frac{1}{x-1}\right)& \text{if }x\ne 1\\ 0& \text{if }x=1\end{array}$
$R{f}^{\prime }\left(1\right)=\underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}$
$=\underset{h\to 0}{\lim }\frac{h\sin \frac{1}{h}-0}{h}=\underset{h\to 0}{\lim }\sin \frac{1}{h}$
= a finite number
Let this finite number be l
$L{f}^{\prime }\left(1\right)=\underset{h\to 0}{\lim }\frac{f\left(1-h\right)-f\left(1\right)}{-h}$
$=\underset{h\to 0}{\lim }\frac{-h\sin \left(\frac{1}{-h}\right)}{-h}$
$=\underset{h\to 0}{\lim }\sin \left(\frac{1}{-h}\right)=-\underset{h\to 0}{\lim }\sin \left(\frac{1}{h}\right)$
= - (a finite number)
Thus Rf'(1) $\ne$ Lf '(1)
$\therefore$ f is not differentiable at x = 1
Also, $f^{\prime }\left(0\right)=\sin \frac{1}{\left(x-1\right)}-\frac{x-1}{{\left(x-1\right)}^2}\cos \left(\frac{1}{x-1}\right){]}_{x=0}$
= -sin 1 + cos 1
$\therefore$ f is differentiable at x = 0