If the function \(f(x) = \begin{cases}\displaystyle\frac{1 - \cos 4x}{x^{2}}, & x \ne 0\\ k, & x = 0\end{cases}\) is continuous at \(x = 0\), then the value of k is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 8
f is continuous at 0 when \(f(0) = \lim_{x \to 0} f(x)\), so k equals this limit.
Using \(1 - \cos 2\theta = 2\sin^{2}\theta\) with \(\theta = 2x\):
\( \displaystyle \frac{1 - \cos 4x}{x^{2}} = \frac{2\sin^{2} 2x}{x^{2}} \)
\( \displaystyle = 2 \times 4 \times \left(\frac{\sin 2x}{2x}\right)^{2} \)
Since \(\lim_{t \to 0}\frac{\sin t}{t} = 1\) (here t = 2x):
\( \displaystyle \lim_{x \to 0}\frac{1 - \cos 4x}{x^{2}} = 8 \times 1^{2} = 8 \)
So k = 8.