1. The problem is to find the limit $$\lim_{x\to 0}\frac{\arcsin(x)}{x}$$. 2. Recall the important limit rule: $$\lim_{x\to 0}\frac{\sin x}{x} = 1$$ and the fact that $$\arcsin(x)$$ is the inverse function of $$\sin x$$. 3. Using the fact that $$\arcsin(x) \approx x$$ when $$x$$ is close to 0 (since $$\arcsin(x)$$ is continuous and differentiable at 0 with derivative 1), we can write: $$\lim_{x\to 0}\frac{\arcsin(x)}{x} = 1$$. 4. This is because the numerator and denominator both approach 0, and their ratio approaches the derivative of $$\arcsin(x)$$ at 0, which is 1. Final answer: $$\boxed{1}$$