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!