1. **State the problem:** We want to find the limit $$\lim_{x \to 1^+} \left( \frac{x}{x-1} - \frac{1}{\ln x} \right).$$\n\n2. **Analyze the behavior near $x=1$: ** As $x \to 1^+$, both denominators $x-1$ and $\ln x$ approach 0, so we expect an indeterminate form.\n\n3. **Use expansions:** Let $x = 1 + h$ with $h \to 0^+$. Then:\n- $x - 1 = h$\n- $\ln x = \ln(1 + h) \approx h - \frac{h^2}{2} + \frac{h^3}{3} + \cdots$\n\n4. **Rewrite the expression:**\n$$\frac{x}{x-1} - \frac{1}{\ln x} = \frac{1 + h}{h} - \frac{1}{h - \frac{h^2}{2} + \cdots}.$$\n\n5. **Simplify each term:**\n- $\frac{1 + h}{h} = \frac{1}{h} + 1$\n- For the second term, use the expansion for $\frac{1}{\ln x}$:\n$$\frac{1}{h - \frac{h^2}{2} + \cdots} = \frac{1}{h} \cdot \frac{1}{1 - \frac{h}{2} + \cdots} \approx \frac{1}{h} \left(1 + \frac{h}{2} + \cdots \right) = \frac{1}{h} + \frac{1}{2} + \cdots$$\n\n6. **Subtract the two terms:**\n$$\left( \frac{1}{h} + 1 \right) - \left( \frac{1}{h} + \frac{1}{2} \right) = 1 - \frac{1}{2} = \frac{1}{2}.$$\n\n7. **Conclusion:** The limit is $$\boxed{\frac{1}{2}}.$$