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.