Question: lim $x\to 0$ $\frac{\sin^{-1}(x)}{8x}$
1. **State the problem:**
Find the limit $$\lim_{x \to 0} \frac{\sin^{-1}(x)}{8x}$$.
2. **Recall the formula and important rule:**
As $x$ approaches 0, $\sin^{-1}(x)$ behaves like $x$ because $\sin^{-1}(x) \approx x$ for small $x$.
3. **Apply the approximation:**
$$\lim_{x \to 0} \frac{\sin^{-1}(x)}{8x} \approx \lim_{x \to 0} \frac{x}{8x}$$
4. **Simplify the fraction:**
$$\lim_{x \to 0} \frac{\cancel{x}}{8\cancel{x}} = \lim_{x \to 0} \frac{1}{8}$$
5. **Evaluate the limit:**
Since the expression simplifies to a constant, the limit is:
$$\frac{1}{8}$$
**Final answer:**
$$\boxed{\frac{1}{8}}$$