Question

The value of the limit \(\lim_{x \to 0} \frac{(2 + \cos 3x) \sin^2 x}{x \tan(2x)}\) is equal to:

• A. \(\frac{3}{2}\)
• B. 2
• C. \(\frac{1}{2}\)
• D. 3
• E. 0

Hint:

Use trigonometric identities and limits for small angles to simplify expressions and find limits effectively, especially when dealing with trigonometric functions.

Answer 1

The Correct Option is A

First, simplify and analyze the limit using trigonometric identities and small-angle approximations: \[ \lim_{x \to 0} \frac{(2 + \cos 3x) \sin^2 x}{x \tan(2x)} \] As \(x \to 0\), \(\cos 3x \approx 1\) and \(\sin x \approx x\), \(\tan 2x \approx 2x\). Substituting these approximations into the limit: \[ = \lim_{x \to 0} \frac{(2 + 1) x^2}{x \cdot 2x} \] \[ = \lim_{x \to 0} \frac{3x^2}{2x^2} \] \[ = \frac{3}{2} \] Thus, the value of the limit is \(\frac{3}{2}\).