Question

Choose the most appropriate option.

\[ \lim_{x \to 0} \frac{\ln \cos 2x}{\sin 2x} \]

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

Hint:

Using the approximation \( \sin x \approx x \) near \( x = 0 \) helps to solve limit problems more efficiently.

Answer 1

The Correct Option is A

Step 1: Direct Substitution

First, substitute \( x = 0 \):

\[ \ln \cos 2x \to \ln 1 = 0 \quad \text{and} \quad \sin 2x \to 0 \]

This gives the indeterminate form \(\frac{0}{0}\), so we apply L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule

Differentiate the numerator and denominator with respect to \( x \):

\[ \text{Numerator: } \frac{d}{dx} [\ln \cos 2x] = \frac{1}{\cos 2x} \cdot (-\sin 2x) \cdot 2 = -2 \tan 2x \]

\[ \text{Denominator: } \frac{d}{dx} [\sin 2x] = 2 \cos 2x \]

The limit becomes:

\[ \lim_{x \to 0} \frac{-2 \tan 2x}{2 \cos 2x} = \lim_{x \to 0} \frac{-\tan 2x}{\cos 2x} \]

Step 3: Evaluate the Simplified Limit

Substitute \( x = 0 \):

\[ \tan 2x \to 0 \quad \text{and} \quad \cos 2x \to 1 \]

Thus:

\[ \lim_{x \to 0} \frac{-\tan 2x}{\cos 2x} = \frac{-0}{1} = 0 \]