1. We are asked to find the limit: $$\lim_{x \to \left(\frac{1}{2}\right)^-} \frac{\ln(1 - 2x)}{\tan(\pi x)}.$$\n\n2. First, note the behavior of the function inside the logarithm and the tangent as $x$ approaches $\frac{1}{2}$ from the left.\n- Inside the logarithm: $1 - 2x$. When $x \to \frac{1}{2}^-$, $1 - 2x \to 0^+$, so $\ln(1 - 2x) \to -\infty$.\n- For the denominator: $\tan(\pi x)$, as $x \to \frac{1}{2}$, $\pi x \to \frac{\pi}{2}$ from the left, and $\tan(\theta)$ approaches $+\infty$ as $\theta \to \frac{\pi}{2}^-$, so $\tan(\pi x) \to +\infty$.\n\n3. The limit is of the form $\frac{-\infty}{+\infty}$, which is an indeterminate form. We can use substitution and L'Hôpital's Rule.\n\n4. Let $t = 1 - 2x$. Then as $x \to \frac{1}{2}^-$, $t \to 0^+$. Also, $x = \frac{1 - t}{2}$.\n\n5. Rewrite the limit in terms of $t$: $$\lim_{t \to 0^+} \frac{\ln(t)}{\tan\left(\pi \frac{1 - t}{2}\right)} = \lim_{t \to 0^+} \frac{\ln(t)}{\tan\left(\frac{\pi}{2} - \frac{\pi t}{2}\right)}.$$\n\n6. Use the identity $\tan\left(\frac{\pi}{2} - \theta\right) = \cot(\theta)$, so denominator becomes $\cot\left(\frac{\pi t}{2}\right)$.\n\n7. Thus, the limit is $$\lim_{t \to 0^+} \frac{\ln(t)}{\cot\left(\frac{\pi t}{2}\right)} = \lim_{t \to 0^+} \ln(t) \cdot \tan\left(\frac{\pi t}{2}\right).$$\n\n8. For small $t$, $\tan\left(\frac{\pi t}{2}\right) \approx \frac{\pi t}{2}$. So the limit approximates to $$\lim_{t \to 0^+} \ln(t) \cdot \frac{\pi t}{2} = \frac{\pi}{2} \lim_{t \to 0^+} t \ln(t).$$\n\n9. It is known that $\lim_{t \to 0^+} t \ln(t) = 0$. Since $t \ln(t)$ approaches $0$ from the negative side, the product tends to $0$.\n\n10. Therefore, the original limit is $$\boxed{0}.$$