The value of k for which the function \(f(x) = \begin{cases}\displaystyle\frac{1 - \cos 8x}{16x^2}, & \text{if } x \ne 0\\ k, & \text{if } x = 0\end{cases}\) is continuous at \(x = 0\) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2
f is continuous at x = 0 when \(k = \lim_{x \to 0} f(x)\). Using \(1 - \cos 8x = 2\sin^2 4x\):
\(\displaystyle \lim_{x \to 0} \frac{1 - \cos 8x}{16x^2} = \lim_{x \to 0} \frac{2\sin^2 4x}{16x^2}\)
\(\displaystyle = \lim_{x \to 0} 2\left(\frac{\sin 4x}{4x}\right)^2\)
Since \(\frac{\sin \theta}{\theta} \to 1\) as \(\theta \to 0\), the limit is 2 × 1 = 2. So k = 2.
Shortcut: By the standard limit
\(\displaystyle \lim_{x \to 0} \frac{1 - \cos ax}{x^2} = \frac{a^2}{2}\)
with a = 8, the limit is
\(\displaystyle \frac{1}{16} \times \frac{64}{2} = 2\)