Find the value of $k$ for which the function $f$ given as $f(x) = \begin{cases} \frac{1 - \cos x}{2x^2}, & \text{if } x \neq 0 \\ k, & \text{if } x = 0 \end{cases} \text{ is continuous at } x = 0$.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $k = \frac{1}{4}$ ##
$f(x) = \begin{cases} \frac{1 - \cos x}{2x^2}, & \text{if } x \neq 0 \\ k, & \text{if } x = 0 \end{cases}$ is continuous at $x = 0$
$\text{LHL} = \lim\limits_{x \to 0^-} f(x) = \lim\limits_{h \to 0} \frac{1 - \cos(0 - h)}{2(0 - h)^2}$
$= \lim\limits_{h \to 0} \frac{1 - \cos h}{2h^2}$
$= \lim\limits_{h \to 0} \frac{2 \sin^2(h/2)}{2h^2}$
$= \lim\limits_{h \to 0} \frac{1}{4} \left( \frac{\sin(h/2)}{2h/2} \right)^2 = \frac{1}{4}$
$f(x)$ is continuous at $x = 0$.
$∴k = \frac{1}{4} \quad \left( ∵f(0) = \lim\limits_{x \to 0} f(x) \right)$