To solve the limit:

\[
\lim_{x \to 0} \frac{\sin(8x)}{\sin(9x)}
\]

We can use the small-angle approximation for sine, which states that \(\sin(x) \approx x\) as \(x\) approaches 0. Applying this approximation to both \(\sin(8x)\) and \(\sin(9x)\):

\[
\lim_{x \to 0} \frac{\sin(8x)}{\sin(9x)} \approx \lim_{x \to 0} \frac{8x}{9x} = \frac{8}{9}
\]

Thus, the limit is:

\[
\lim_{x \to 0} \frac{\sin(8x)}{\sin(9x)} = \frac{8}{9}
\]

Would you like more details or have any questions?

Here are 5 related questions to explore this concept further:
1. How would the result change if we had \(\lim_{x \to 0} \frac{\sin(5x)}{\sin(3x)}\)?
2. Can you use L'Hopital's Rule to verify the result?
3. How does the small-angle approximation work, and when can it be used?
4. What would happen if we took the limit of \(\frac{\sin(kx)}{\sin(mx)}\) as \(x \to 0\) for any constants \(k\) and \(m\)?
5. What is the limit of \(\lim_{x \to 0} \frac{\tan(8x)}{\tan(9x)}\)?

**Tip**: The small-angle approximation is useful for simplifying trigonometric limits as \(x \to 0\), especially when sine or tangent functions are involved.

Explore the limit of sin(8x)/sin(9x) as x approaches 0 using the small-angle approximation. Learn the step-by-step solution, including how to apply trigonometric concepts and limit formulas. Suitable for students in Grades 10-12.

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