The limit problem presented in the image is:

$\lim_{{x \to 0}} \frac{\sin(8x)}{\tan(9x)}$

We can solve this using trigonometric limit rules. One useful rule is:

$\lim_{{x \to 0}} \frac{\sin(kx)}{kx} = 1 \quad \text{and} \quad \lim_{{x \to 0}} \frac{\tan(kx)}{kx} = 1$

We can rewrite the given limit by factoring constants from both the sine and tangent terms:

$\lim_{{x \to 0}} \frac{\sin(8x)}{\tan(9x)} = \lim_{{x \to 0}} \frac{\frac{\sin(8x)}{x}}{\frac{\tan(9x)}{x}}$

Using the trigonometric limits:

$\lim_{{x \to 0}} \frac{\sin(8x)}{x} = 8 \quad \text{and} \quad \lim_{{x \to 0}} \frac{\tan(9x)}{x} = 9$

Thus, the limit becomes:

$\frac{8}{9}$

So the solution is:

$\lim_{{x \to 0}} \frac{\sin(8x)}{\tan(9x)} = \frac{8}{9}$

Would you like further details or clarification? Here are some related questions:

1. What is the general form of the trigonometric limit rule?
2. How can this approach be applied to other trigonometric limit problems?
3. Why are small-angle approximations useful in limit calculations?
4. Can you explain why $\lim_{{x \to 0}} \frac{\sin(kx)}{kx} = 1$?
5. What happens if different coefficients are used for $x$ in sine and tangent?

Tip: Recognizing standard limits such as $\lim_{{x \to 0}} \frac{\sin(x)}{x} = 1$ greatly simplifies limit problems involving trigonometric functions.