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The limit to be evaluated is:

\[
\lim_{x \to 0} \frac{\csc x - \cos x \cdot \cot x}{2x}
\]

We will solve this by simplifying the expression and using known limit properties.

1. **Recall trigonometric identities:**
   - \( \csc x = \frac{1}{\sin x} \)
   - \( \cot x = \frac{\cos x}{\sin x} \)

Using these, the expression becomes:

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

Now, recall that \( 1 - \cos^2 x = \sin^2 x \), so the expression simplifies to:

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

2. **Evaluate the limit:**

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

We know from standard limits that \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). Therefore, we have:

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

Thus, the value of the limit is:

\[
\boxed{\frac{1}{2}}
\]

Would you like further clarification, or do you have any additional questions?

Here are 5 related questions:
1. How does the identity \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) arise?
2. What are other common trigonometric limits that are useful in calculus?
3. Can you explain how L'Hopital's Rule applies to limits like this?
4. How do you derive the Taylor expansion for \( \sin x \) and \( \cos x \)?
5. What happens to the limit if we change \( 2x \) in the denominator to another function?

**Tip:** Memorizing standard trigonometric limits such as \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) and \( \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \) can significantly simplify solving these types of problems.

Find the limit lim (x→0) [(csc x - cos x cot x) / 2x]

Learn how to find the limit of the expression (csc x - cos x cot x) / 2x as x approaches 0. This problem uses key trigonometric identities and limit theorems to simplify the expression and reach the solution. A step-by-step solution is provided.

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