Question

The value of \(\underset{x\rightarrow 0}{lim}(\frac{8}{x^8})[1-cos\frac{x^2}{2}-cos\frac{x^2}{4}+cos\frac{x^2}{2}\cdot cos\frac{x^2}{4}]\) is equal to ...

• A. \(\frac{1}{8}\)
• B. \(\frac{1}{32}\)
• C. \(\frac{1}{16}\)
• D. \(0\)

Hint:

The bracket factorises as (1 - cos A)(1 - cos B).

Answer 1

The Correct Option is B

Step 1: Half-Angle Form:
Use $1-\cos\theta=2\sin^2(\theta/2)$. Then $1-\cos A=2\sin^2\dfrac{x^2}{4}$ and $1-\cos B=2\sin^2\dfrac{x^2}{8}$.

Step 2: Product:
Bracket $=4\sin^2\dfrac{x^2}{4}\sin^2\dfrac{x^2}{8}$. As $x\to0$, $\sin\theta\sim\theta$, so this is $4\cdot\dfrac{x^4}{16}\cdot\dfrac{x^4}{64}=\dfrac{x^8}{256}$.

Step 3: Limit:
Multiply by $8/x^8$: $\dfrac{8}{256}=\dfrac1{32}$. Option (B).

Final Answer:
Option (B). \[ \boxed{\text{(B) } \frac{1}{32}} \]