1. **State the problem:** We want to find the limit $$\lim_{x \to 0^+} \ln\left(\frac{x}{\sin x}\right).$$\n\n2. **Recall important facts:** As $x \to 0$, $\sin x \approx x - \frac{x^3}{6} + \cdots$, so $\frac{x}{\sin x} \to 1$. The natural logarithm function $\ln y$ is continuous near $y=1$.\n\n3. **Rewrite the limit inside the logarithm:** $$\frac{x}{\sin x} = \frac{x}{x - \frac{x^3}{6} + \cdots} = \frac{x}{x\left(1 - \frac{x^2}{6} + \cdots\right)} = \frac{1}{1 - \frac{x^2}{6} + \cdots}.$$\n\n4. **Simplify the fraction:** $$\frac{1}{1 - \frac{x^2}{6} + \cdots} = 1 + \frac{x^2}{6} + \cdots$$ using the expansion for $\frac{1}{1 - y} = 1 + y + y^2 + \cdots$ for small $y$.\n\n5. **Apply the logarithm expansion:** For $z$ close to 0, $\ln(1+z) \approx z$. Here, $z = \frac{x^2}{6} + \cdots$. So, $$\ln\left(\frac{x}{\sin x}\right) = \ln\left(1 + \frac{x^2}{6} + \cdots\right) \approx \frac{x^2}{6} + \cdots.$$\n\n6. **Evaluate the limit:** As $x \to 0^+$, $\frac{x^2}{6} \to 0$, so $$\lim_{x \to 0^+} \ln\left(\frac{x}{\sin x}\right) = 0.$$\n\n**Final answer:** $$\boxed{0}.$$