Examine the continuity of the function $f(x) = \begin{cases} \frac{1 - \cos 2x}{x^2}, & \text{if } x \neq 0 \\ 5, & \text{if } x = 0 \end{cases}$ at $x = 0$.

Discontinuous at $x = 0$ because $\lim\limits_{x \to 0} f(x) = 2$, which is not equal to $f(0)$.

The correct answer is Option (2) → Discontinuous at $x = 0$ because $\lim\limits_{x \to 0} f(x) = 2$, which is not equal to $f(0)$. ##

We have, $f(x) = \begin{cases} \frac{1 - \cos 2x}{x^2}, & \text{if } x \neq 0 \\ 5, & \text{if } x = 0 \end{cases}$ at $x = 0$.

At $x = 0$, $\text{LHL} = \lim\limits_{x \to 0^-} \frac{1 - \cos 2x}{x^2}$

Put $x = 0 - h$, $= \lim\limits_{h \to 0} \frac{1 - \cos 2(0 - h)}{(0 - h)^2}$

$= \lim\limits_{h \to 0} \frac{1 - \cos 2h}{h^2} \quad [∵\cos(-\theta) = \cos \theta]$

$= \lim\limits_{h \to 0} \frac{1 - (1 - 2 \sin^2 h)}{h^2} \quad [∵\cos 2\theta = 1 - 2 \sin^2 \theta]$

$= \lim\limits_{h \to 0} \frac{2(\sin h)^2}{(h)^2} \quad \left[ ∵\lim\limits_{h \to 0} \frac{\sin h}{h} = 1 \right]$

$= 2$

$\text{RHL} = \lim\limits_{x \to 0^+} \frac{1 - \cos 2x}{x^2}$

Put $x = 0 + h$, $= \lim\limits_{h \to 0} \frac{1 - \cos 2(0 + h)}{(0 + h)^2}$

$= \lim\limits_{h \to 0} \frac{2 \sin^2 h}{h^2} = 2 \quad \left[ ∵\lim\limits_{h \to 0} \frac{\sin h}{h} = 1 \right]$

and $f(0) = 5$

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

Hence, $f(x)$ is not continuous at $x = 0$.