Question

Let $f(x) = \begin{cases} (x-1) \sin \frac{1}{x-1} & \quad \text{if } x \neq 1\\ 0 & \quad \text{if }x =1 \end{cases} $
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{cases} (x-1) \sin \left(\frac{1}{x-1} \right) & \quad \text{if } x \neq 1\\ 0 & \quad \text{if }x =1 \end{cases} $
$Rf' \left(1\right) = \displaystyle\lim_{h \to0} \frac{f\left(1+h\right)-f\left(1\right)}{h}$
$ = \displaystyle\lim_{h\to0} \frac{h \sin \frac{1}{h} - 0}{h} = \displaystyle\lim_{h \to0} \sin \frac{1}{h}$
= a finite number
Let this finite number be l
$ L f'\left(1\right) = \displaystyle\lim_{h \to0} \frac{f\left(1-h\right)-f\left(1\right)}{-h}$
$= \displaystyle\lim_{h \to0} \frac{-h \sin\left(\frac{1}{-h}\right)}{-h} $
$= \displaystyle\lim_{h \to0} \sin\left(\frac{1}{-h}\right) = - \displaystyle\lim_{h \to0} \sin\left(\frac{1}{h}\right) $
= - (a finite number)
Thus Rf'(1) $\neq$ Lf '(1)
$\therefore $ f is not differentiable at x = 1
Also, $f'\left(0\right) = \sin \frac{1}{\left(x-1\right)}- \frac{x-1}{\left(x-1\right)^{2}} \cos\left(\frac{1}{x-1}\right) \bigg]_{x =0}$
= -sin 1 + cos 1
$\therefore $ f is differentiable at x = 0