The given limit is:

$\lim_{{x \to 0}} \frac{{\sin(10x)}}{x}$

To solve this, we can use a well-known limit result:

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

Here, $k = 10$. Therefore:

$\lim_{{x \to 0}} \frac{{\sin(10x)}}{x} = 10$

The limit evaluates to $10$.

Would you like more details on the steps or have any other questions?

Here are some related questions:

1. How does this formula for the limit $\frac{\sin(kx)}{x}$ derive from the small-angle approximation?
2. Can this technique be applied to other trigonometric functions?
3. What happens if we change $\sin(10x)$ to $\cos(10x)$?
4. What if we approach this using L'Hôpital's Rule?
5. How can this limit help us understand the derivative of the sine function?

Tip: Memorizing the standard limit $\lim_{{x \to 0}} \frac{\sin(x)}{x} = 1$ can simplify many trigonometric limit problems.