1. **State the problem:** Find the limit $$\lim_{x \to 0} (1-2x)^{\frac{1}{x}}.$$\n\n2. **Recall the formula:** Limits of the form $$\lim_{x \to 0} (1 + ax)^{\frac{1}{x}} = e^a$$ are common and useful. Here, the base is $$1 - 2x$$, so $$a = -2$$.\n\n3. **Rewrite the expression:** We can write the limit as $$\lim_{x \to 0} \exp\left( \frac{1}{x} \ln(1 - 2x) \right).$$\n\n4. **Use the expansion for $$\ln(1 + u)$$:** For small $$u$$, $$\ln(1 + u) \approx u - \frac{u^2}{2} + \cdots$$. Here, $$u = -2x$$, so $$\ln(1 - 2x) \approx -2x - \frac{(-2x)^2}{2} = -2x - \frac{4x^2}{2} = -2x - 2x^2.$$\n\n5. **Substitute back:** $$\frac{1}{x} \ln(1 - 2x) \approx \frac{1}{x} (-2x - 2x^2) = -2 - 2x.$$\n\n6. **Take the limit as $$x \to 0$$:** $$\lim_{x \to 0} (-2 - 2x) = -2.$$\n\n7. **Therefore, the original limit is:** $$\lim_{x \to 0} (1-2x)^{\frac{1}{x}} = \lim_{x \to 0} \exp\left( \frac{1}{x} \ln(1 - 2x) \right) = e^{-2}.$$\n\n**Final answer:** $$e^{-2}.$$