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!