Question

The limit \( \lim_{x \to 0} \frac{3 \sin^2 2x}{x^2} \) is equal to:

• A. 3
• B. 2
• C. 6
• D. \( \frac{3}{2} \)
• E. 12

Hint:

Apply the standard limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) to simplify the expression.

Answer 1

Answer: (E)

We are given \( \lim_{x \to 0} \frac{3 \sin^2 2x}{x^2} \). 
Using the standard limit result \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), we can simplify the expression as: \[ \lim_{x \to 0} \frac{3 \sin^2 2x}{x^2} = 3 \cdot \lim_{x \to 0} \frac{\sin^2 2x}{x^2} = 3 \cdot \lim_{x \to 0} \left( \frac{\sin 2x}{x} \right)^2. \] Since \( \lim_{x \to 0} \frac{\sin 2x}{x} = 2 \), we get: \[ \lim_{x \to 0} \frac{3 \sin^2 2x}{x^2} = 3 \cdot 2^2 = 12. \] Thus, the limit is 12.