To evaluate the limit

$\lim_{x \to 0} \frac{\sin^{100}(x)}{x^{99} \sin(2x)},$

let's break it down step by step.

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Step 1: Simplify the powers of $\sin(x)$

Using the small-angle approximation, $\sin(x) \approx x$ as $x \to 0$. Therefore: $\sin^{100}(x) \approx x^{100}.$

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Step 2: Simplify $\sin(2x)$

Using the small-angle approximation again, $\sin(2x) \approx 2x$ as $x \to 0$.

Thus: $\sin(2x) \approx 2x.$

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Step 3: Substitute approximations into the limit

Substituting these into the limit, we get: $\lim_{x \to 0} \frac{\sin^{100}(x)}{x^{99} \sin(2x)} \approx \lim_{x \to 0} \frac{x^{100}}{x^{99} \cdot 2x}.$

Simplify the numerator and denominator: $\frac{x^{100}}{x^{99} \cdot 2x} = \frac{x^{100}}{2x^{100}} = \frac{1}{2}.$

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Step 4: Final result

The limit evaluates to: $\boxed{\frac{1}{2}}$

Let me know if you have questions or need clarification! 😊

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Related Questions:

1. How do small-angle approximations help simplify trigonometric limits?
2. What is the derivation of $\sin(x) \approx x$ for small $x$?
3. How can we use L'Hôpital's Rule to solve this limit?
4. What are common strategies to evaluate limits involving trigonometric functions?
5. How does the approximation $\sin(2x) \approx 2x$ compare to Taylor series expansion?

Tip: For limits involving trigonometric functions near 0, remember small-angle approximations!