Find the value(s) of $k$ so that the following function is continuous at $x = 0$. $f(x) = \begin{cases} \frac{1 - \cos kx}{x \sin x}, & \text{if } x \neq 0 \\ \frac{1}{2}, & \text{if } x = 0 \end{cases}$
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $k = \pm 1$ ##
$\lim\limits_{x \to 0} \frac{1 - \cos kx}{x \sin x} = \lim\limits_{x \to 0} \frac{2 \sin^2 \left( \frac{kx}{2} \right)}{x \sin x}$
$= \lim\limits_{x \to 0} \frac{\frac{2 \sin^2 \left( \frac{kx}{2} \right)}{x^2}}{\frac{x \sin x}{x^2}}$
$= \frac{\lim\limits_{x \to 0} \frac{2 \sin^2 \left( \frac{kx}{2} \right)}{\left( \frac{kx}{2} \right)^2} \times \left( \frac{k}{2} \right)^2}{\lim\limits_{x \to 0} \frac{\sin x}{x}}$
$= \frac{2 \times 1 \times \frac{k^2}{4}}{1}$
$∵f(x)$ is continuous at $x = 0$,
$∴\lim\limits_{x \to 0} f(x) = f(0)$
$⇒\frac{k^2}{2} = \frac{1}{2}$
$⇒ k^2 = 1$
$⇒k = \pm 1$