The value of $k$ for which the function $f(x) =\left\{\begin{matrix}\frac{1-\cos 8x}{16x^2},&\text{if x≠0}\\k,&\text{if x=0}\end{matrix}\right.$ is continuous at $x=0$ is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2
k = $\lim_{x \to 0} \frac{1 - \cos 8x}{16x^2}$
Since numerator and denominator $\to 0$ as $x \to 0$, apply L’Hôpital’s Rule:
Derivative of numerator = $8 \sin 8x$
Derivative of denominator = $32x$
So,
$k = \lim_{x \to 0} \frac{8 \sin 8x}{32x}$
$= \frac{1}{4} \times \lim_{x \to 0} \frac{\sin 8x}{x}$
$= \frac{1}{4} \times 8$
$= 2$