To find the limit of \( \frac{2x}{\sin(x)} \) as \( x \to 0 \), we can approach this using standard limit results and analysis:

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

### Step 1: Use a known limit

We know the well-established limit:

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

This implies that as \( x \to 0 \), \( \sin(x) \) behaves similarly to \( x \). Therefore, we can rewrite the expression as:

\[
\frac{2x}{\sin(x)} = 2 \cdot \frac{x}{\sin(x)}
\]

### Step 2: Apply the limit

Now, using the known limit:

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

Thus:

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

### Final Answer:

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

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Would you like more explanation on this, or do you have any other questions?

### Related Questions:
1. What happens to \( \frac{x}{\sin(x)} \) as \( x \to 0 \)?
2. How can we use L'Hopital's Rule to evaluate limits like this?
3. What is the Taylor series expansion for \( \sin(x) \) around \( x = 0 \)?
4. How does the limit of \( \frac{x}{\sin(x)} \) relate to the derivative of \( \sin(x) \)?
5. What are some other limits involving trigonometric functions?

**Tip:** Always try to use standard limit results like \( \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \) for simplifying expressions involving trigonometric functions near 0!

Find the limit of 2x/sin(x) as x approaches 0 with a detailed explanation. Learn how to use standard limit results, such as lim(x -> 0) sin(x)/x = 1, to solve this trigonometric limit problem. Suitable for high school students in Grades 11-12.

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