Question

The value of \(\displaystyle \lim_{x\to 0}\left[\frac{\cos 2x-\cos 4x}{x^2}\right]\) is __________________.

Hint:

- For small \(x\), use \(\sin x \sim x\) or standard limits \(\lim_{x\to0}\frac{\sin x}{x}=1\).
- Sum-to-product identities simplify trig limits quickly.

Answer 1

Correct Answer: 6

Step 1: Use the identity \(\cos A-\cos B=-2\sin\!\left(\tfrac{A+B}{2}\right)\sin\!\left(\tfrac{A-B}{2}\right)\).
\(\cos 2x-\cos 4x=-2\sin(3x)\sin(-x)=2\sin(3x)\sin x\).
Step 2: Hence \[\frac{\cos 2x-\cos 4x}{x^2}=2\cdot\frac{\sin(3x)}{x}\cdot\frac{\sin x}{x}.\] As \(x\to0\), \(\frac{\sin(3x)}{x}\to 3\) and \(\frac{\sin x}{x}\to1\).
Step 3: Therefore, the limit \(=2\cdot 3 \cdot 1=6\).