Let f ( x ) = { ( x - 1 ) sin ( 1 x - 1 ) , if x 1 0 , if x = 1 . Then, which of the following is true?

Question

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

Smart Achievers · Accepted Answer

(4)

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

Explanation:

We observe that,

$\lim_{x \to 1} \frac{f (x) - f (1)}{x - 1} = \lim_{x \to 1} \frac{(x - 1) \sin (\frac{1}{x - 1})}{x - 1}$

$= \lim_{x \to 1} \sin (\frac{1}{x - 1})$

= An oscillating number between -1 and 1

$∴$ $\lim_{x \to 1} \frac{f (x) - f (1)}{x - 1}$ does not exist.

$\Rightarrow$ $f (x)$ is not differentiable at $x = 1$ .

$= \lim_{x \to 0} \frac{f (x) - f (0)}{x - 0}$

$= \lim_{x \to 0} \frac{(x - 1) \sin (\frac{1}{x - 1}) - \sin 1}{x}$

$= \lim_{x \to 0} \frac{x \sin (\frac{1}{x - 1})}{x} - \lim_{x \to 0} \frac{\sin (\frac{1}{x - 1}) + \sin 1}{x}$

$= - \sin 1 - \lim_{x \to 0} \frac{2 \sin (\frac{x}{2 (x - 1)}) \cos {\frac{2 - x}{2 (x - 1)}}}{{\frac{x}{2 (x - 1)}} 2 (x - 1)}$

$= - \sin 1 + \cos 1$

$∴$ $f (x) is differentiable at x = 0$

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