To solve this limit using L'Hospital's Rule, we first need to express it in a form suitable for L'Hospital's Rule. The given limit is of an indeterminate form $1^\infty$, so let's proceed with the following steps:

1. Transform the Expression:

   Consider $L = \lim_{x \to 0} (1 - 2x)^{1/x}.$ Take the natural logarithm of both sides to bring the exponent down: $\ln L = \lim_{x \to 0} \frac{\ln(1 - 2x)}{x}.$

2. Apply L'Hospital's Rule:

   The form $\frac{\ln(1-2x)}{x}$ is of type $\frac{0}{0}$ as $x \to 0$. Differentiate the numerator and denominator: $\text{Numerator: } \frac{d}{dx}[\ln(1-2x)] = -\frac{2}{1-2x}.$ $\text{Denominator: } \frac{d}{dx}[x] = 1.$

   Apply L'Hospital's Rule: $\ln L = \lim_{x \to 0} \frac{-2}{1-2x}.$

3. Evaluate the Limit:

   As $x \to 0$, $\ln L = \frac{-2}{1 - 0} = -2.$

4. Exponentiate to Solve for $L$:

   Since $\ln L = -2$, then $L = e^{-2}.$

So, the original limit is: $\lim_{x \to 0} (1 - 2x)^{1/x} = e^{-2}.$