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: Factorise the Bracket:
Let \(A=\dfrac{x^2}{2}\) and \(B=\dfrac{x^2}{4}\). The bracket is \(1-\cos A-\cos B+\cos A\cos B=(1-\cos A)(1-\cos B)\).

Step 2: Use the Small-Angle Estimate:
For small \(\theta\), \(1-\cos\theta\approx\dfrac{\theta^2}{2}\). So
\[ 1-\cos A\approx\frac{x^4}{8},\qquad 1-\cos B\approx\frac{x^4}{32} \]

Step 3: Multiply:
The bracket is about \(\dfrac{x^4}{8}\cdot\dfrac{x^4}{32}=\dfrac{x^8}{256}\).

Step 4: Take the Limit:
\[ \lim_{x\to0}\frac{8}{x^8}\cdot\frac{x^8}{256}=\frac{8}{256}=\frac{1}{32} \]
Option (A) \(\tfrac18\) would need the bracket to be \(x^8/64\), and option (D) 0 would need a higher power of x. So (B) is correct.

Final Answer:
The limit is \(\dfrac1{32}\), option (B). \[ \boxed{\text{(B) } \frac{1}{32}} \]