The value of ‘\( \lambda \)’ for which the function \( f(x)=\begin{cases}\frac{1-\cos 4x}{4x^2} & ,if\quad x\ne 0\\ \lambda & ,if\quad x=0\end{cases} \) is continuous at \( x=0 \), is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2
For continuity at 0, \( \lambda=\lim_{x\to 0}f(x) \). Since \( 1-\cos 4x=2\sin^2 2x \):
\( \displaystyle \lim_{x\to 0}\frac{1-\cos 4x}{4x^2}=\lim_{x\to 0}\frac{2\sin^2 2x}{4x^2} \)
\( \displaystyle =\lim_{x\to 0}2\left(\frac{\sin 2x}{2x}\right)^2 \)
\( \displaystyle =2\times 1=2 \)
So \( \lambda=2 \).