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We are asked to solve the following limit problem as \( x \to 0 \):

\[
\lim_{x \to 0} \frac{1 - \cos x}{2x \cdot \sin x}
\]

### Step 1: Use trigonometric limit identities
The following two important limit identities are useful for solving this type of problem:
- \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)
- \( \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \)

### Step 2: Simplifying the expression
We can rewrite the limit as:

\[
\frac{1 - \cos x}{2x \sin x} = \frac{1 - \cos x}{x^2} \cdot \frac{x}{2 \sin x}
\]

Now, apply the known limits:
- \( \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \)
- \( \lim_{x \to 0} \frac{x}{\sin x} = 1 \)

Thus, the limit becomes:

\[
\lim_{x \to 0} \frac{1 - \cos x}{2x \sin x} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}
\]

### Conclusion:
The value of the limit is \( \frac{1}{4} \).

The correct answer is \( \frac{1}{4} \).

---

Would you like further explanation or details on any of the steps? Here are some additional questions you might find useful:

1. What other trigonometric limits should I be familiar with?
2. How can L'Hopital's rule be applied to limits like this?
3. How is the small-angle approximation used in solving trigonometric limits?
4. Can you explain why the identity \( \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \) holds?
5. What other problems can be solved using these trigonometric limits?

**Tip:** Always remember the small-angle approximations for sine and cosine when solving limits involving trigonometric functions!

Nilai dari lim x -> 0 (1 - cos(x)) / (2x * sin(x)) = ...

This page explains how to solve the limit lim x -> 0 (1 - cos(x)) / (2x * sin(x)). It uses key trigonometric limits such as lim x -> 0 (sin(x) / x) = 1 and lim x -> 0 ((1 - cos(x)) / x^2) = 1/2. This problem is suitable for Grade 12 students preparing for calculus exams.

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