The expression in the image appears to be:

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

To solve this limit, we can use the standard limit property:

$\lim_{{x \to 0}} \frac{\sin(kx)}{x} = k$

where $k$ is a constant. In this case, $k = 8$. Therefore:

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

So, the result is $8$.

Would you like more details, or do you have any questions?

Here are some related questions to further explore this concept:

1. How would the limit change if the angle were $\sin(5x)$ instead of $\sin(8x)$?
2. Can we evaluate $\lim_{{x \to 0}} \frac{\sin(3x)}{\sin(2x)}$?
3. How would we handle $\lim_{{x \to 0}} \frac{\tan(8x)}{x}$?
4. What happens if we evaluate $\lim_{{x \to 0}} \frac{\sin(8x)}{x^2}$?
5. Can we use L'Hôpital's Rule to solve this limit?

Tip: For limits involving trigonometric functions like $\sin(kx)$ or $\tan(kx)$ as $x \to 0$, remember that the limit $\frac{\sin(kx)}{x} \to k$ simplifies calculations significantly.