Question

Evaluate the limit: \[ \lim_{x \to 0} \frac{1 - \cos x \cos 2x}{\sin^2 x} \]

• A. \( \frac{11}{4} \)
• B. \( \frac{5}{2} \)
• C. \( 3 \)
• D. \( 5 \)

Hint:

For small-angle limit problems, use the standard approximations \( \cos x \approx 1 - \frac{x^2}{2} \) and \( \sin x \approx x \).

Answer 1

The Correct Option is B

Step 1: Expanding trigonometric functions Using approximations: \[ \cos x \approx 1 - \frac{x^2}{2}, \quad \cos 2x \approx 1 - 2x^2 \] \[ \cos x \cos 2x \approx (1 - \frac{x^2}{2})(1 - 2x^2) \] \[ \approx 1 - \frac{x^2}{2} - 2x^2 + O(x^4) \] \[ \approx 1 - \frac{5x^2}{2} \] \[ 1 - \cos x \cos 2x \approx \frac{5x^2}{2} \] Step 2: Evaluating the limit \[ \lim_{x \to 0} \frac{\frac{5x^2}{2}}{x^2} \] \[ = \frac{5}{2} \]