Examine the continuity of the function f(x) = […] at x = a.

The correct answer is Option (1) → Continuous at $x = a$ because $\lim\limits_{x \to a} f(x) = f(a) = 0$. ##

We have,

$f(x) = \begin{cases} |x - a| \sin \frac{1}{x - a}, & \text{if } x \neq a \\ 0, & \text{if } x = a \end{cases} \text{ at } x = a$

At $x = a$,

$\text{LHL} = \lim\limits_{x \to a^-} |x - a| \sin \frac{1}{x - a}$

Put $x = a - h$,

$= \lim\limits_{h \to 0} |a - h - a| \sin \left( \frac{1}{a - h - a} \right)$

$= \lim\limits_{h \to 0} -h \sin \left( \frac{1}{h} \right) \quad [∵\sin(-\theta) = -\sin \theta]$

$= 0 \times [\text{an oscillating number between } -1 \text{ and } 1]$

$= 0$

$\text{RHL} = \lim\limits_{x \to a^+} |x - a| \sin \left( \frac{1}{x - a} \right)$

Put $x = a + h$,

$= \lim\limits_{h \to 0} |a + h - a| \sin \left( \frac{1}{a + h - a} \right)$

$= \lim\limits_{h \to 0} h \sin \frac{1}{h}$

$= 0 \times [\text{an oscillating number between } -1 \text{ and } 1]$

$= 0$

And $f(a) = 0$ [given]

$∴\text{LHL} = \text{RHL} = f(a)$

So, $f(x)$ is continuous at $x = a$.