The limit of \(\sin(x)\) as \(x\) approaches 0 is given by:

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

This is because the sine function is continuous at \(x = 0\), and \(\sin(0)\) equals 0.

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Here are five related questions:

1. What is the limit of \(\cos(x)\) as \(x\) approaches 0?
2. How do you evaluate the limit \(\lim_{{x \to 0}} \frac{\sin(x)}{x}\)?
3. What is the derivative of \(\sin(x)\) at \(x = 0\)?
4. How does the small-angle approximation \(\sin(x) \approx x\) for \(x\) near 0 relate to limits?
5. What is the Taylor series expansion of \(\sin(x)\) around \(x = 0\)?

**Tip:** For very small angles (in radians), \(\sin(x)\) is approximately equal to \(x\), which is useful for simplifying expressions in calculus.

Understand how to evaluate the limit of sin(x) as x approaches 0 with a step-by-step explanation. This problem involves trigonometric limits and uses the squeeze theorem to prove the result. Suitable for advanced high school students and beyond.

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