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We are given the limit equation:

\[
\lim_{x \to 0} \frac{\sin(2x)}{2x} = 1
\]

And we are asked to find:

\[
\lim_{x \to 0} \frac{\tan(2x)}{6x \sec(3x)}
\]

### Step-by-step solution:

1. **Break down the expression**:  
   The expression to evaluate is:

   \[
   \frac{\tan(2x)}{6x \sec(3x)}
   \]

2. **Use approximations for small \( x \)**:  
   As \( x \to 0 \), we can use the small angle approximations for trigonometric functions:
   - \( \tan(2x) \approx 2x \) when \( x \to 0 \)
   - \( \sec(3x) \approx 1 \) because \( \sec(0) = 1 \)

   Therefore, the expression becomes:

   \[
   \frac{2x}{6x \cdot 1} = \frac{2x}{6x} = \frac{2}{6} = \frac{1}{3}
   \]

3. **Final answer**:  
   The limit is \( \frac{1}{3} \), so the correct choice is:

   **Answer: C. \( \frac{1}{3} \)**

Would you like any further details or explanations?

Here are 5 related questions:
1. How do we apply small angle approximations for trigonometric functions?
2. What is the significance of using limits in calculus?
3. Can you explain the behavior of the secant function near zero?
4. How would the limit change if the angle in the tangent were different?
5. What happens if we remove the secant function from this limit expression?

**Tip**: Remember that small angle approximations are helpful when dealing with trigonometric limits as \( x \to 0 \).

It is known that \( \lim_{x \to 0} \frac{\sin(2x)}{2x} = 1 \). What is \( \lim_{x \to 0} \frac{\tan(2x)}{6x \sec(3x)} \)?

Learn how to evaluate the limit of \( \lim_{x \to 0} \frac{\tan(2x)}{6x \sec(3x)} \) using small angle approximations and basic limit theorems. This problem is suitable for high school calculus students and involves key concepts in trigonometry and limits.

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