Examine the continuity of the function $f(x) = \begin{cases} |x| \cos \frac{1}{x}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}$ at $x = 0$.

Continuous at $x = 0$ because $\lim\limits_{x \to 0} f(x) = f(0) = 0$.

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

We have,

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

At $x = 0$,

$\text{LHL} = \lim\limits_{x \to 0^-} |x| \cos \frac{1}{x} = \lim\limits_{h \to 0} |0 - h| \cos \frac{1}{0 - h}$

Put $x = 0 - h$,

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

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

$\text{RHL} = \lim\limits_{x \to 0^+} |x| \cos \frac{1}{x}$

$= \lim\limits_{h \to 0} |0 + h| \cos \frac{1}{(0 + h)}$

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

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

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

Since, $\text{LHL} = \text{RHL} = f(0)$

So, $f(x)$ is continuous at $x = 0$.