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The given problem asks to determine the limit:

\[
\lim_{x \to 0} \frac{7 \sin \left( \frac{1}{3} x \right)}{\tan \left( \frac{1}{2} x \right)}.
\]

### Step-by-step solution:

1. **Use small-angle approximations:**
   For small values of \(x\), the following approximations hold:
   - \(\sin x \approx x\),
   - \(\tan x \approx x\).

2. **Apply the approximations:**
   As \(x \to 0\), we can approximate \(\sin\left( \frac{1}{3} x \right)\) and \(\tan\left( \frac{1}{2} x \right)\):
   \[
   \sin\left( \frac{1}{3} x \right) \approx \frac{1}{3} x \quad \text{and} \quad \tan\left( \frac{1}{2} x \right) \approx \frac{1}{2} x.
   \]

3. **Substitute these into the limit expression:**
   \[
   \lim_{x \to 0} \frac{7 \sin \left( \frac{1}{3} x \right)}{\tan \left( \frac{1}{2} x \right)} = \lim_{x \to 0} \frac{7 \cdot \frac{1}{3} x}{\frac{1}{2} x}.
   \]

4. **Simplify the expression:**
   The \(x\) terms cancel out:
   \[
   = 7 \cdot \frac{\frac{1}{3}}{\frac{1}{2}} = 7 \cdot \frac{2}{3} = \frac{14}{3}.
   \]

Thus, the value of the limit is \( \frac{14}{3} = 4 \frac{2}{3} \).

### Final answer:
The correct answer is \(4 \frac{2}{3}\), which corresponds to the first option.

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Would you like further details on any part of the solution? Here are some related questions you could explore:

1. How can small-angle approximations be derived?
2. What are the conditions for using approximations in limit problems?
3. Can L'Hopital's Rule be used in solving this type of limit?
4. How do limits behave for trigonometric functions as \(x \to 0\)?
5. What other methods exist to simplify limits involving trigonometric functions?

**Tip:** For problems involving trigonometric limits, small-angle approximations are often very useful for quick solutions.

Tentukan nilai dari \( \lim_{x \to 0} \frac{7 \sin\left( \frac{1}{3} x \right)}{\tan\left( \frac{1}{2} x \right)} \).

This problem involves finding the limit of \( \lim_{x \to 0} \frac{7 \sin\left( \frac{1}{3} x \right)}{\tan\left( \frac{1}{2} x \right)} \). Using small-angle approximations for trigonometric functions, we simplify the expression and find the limit value. This solution is suitable for high school students in Grades 10-12.

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