For $x$ close to $0$, $1-2x$ is positive. So $(1-2x)^{1/x} = e^{\ln(1-2x)/x}.$ Since the exponential function is continuous, $\lim_{x\to 0} e^{\ln(1-2x)/x} = e^{\scriptstyle\left(\lim\limits_{x\to 0}\ln(1-2x)/x\right)}$ provided the latter limit exists. So this lets you change the original problem into the problem of determining whether $\lim_{x\to 0}\frac{\ln(1-2x)}{x}$ exists, and if so what the limit is.
(Alternatively, since $\ln$ is continuous, $\lim_{x\to 0}\ln\left((1-2x)^{1/x}\right) = \ln\left(\lim_{x\to 0}(1-2x)^{1/x}\right)$ so you can do the limit of the natural log instead).
Now, the limit $\lim_{x\to 0}\frac{\ln(1-2x)}{x}$ is an indeterminate of type $\frac{0}{0}$, so you can try using L'Hopital's rule. We get \begin{align*} \lim_{x\to 0}\frac{\ln(1-2x)}{x} &= \lim_{x\to 0}\frac{(\ln(1-2x))'}{x'} &\text{(L'Hopital's Rule)}\\ &= \lim_{x\to 0}\frac{\quad\frac{1}{1-2x}(1-2x)'\quad}{1}\\ &= \lim_{x\to 0}\frac{(1-2x)'}{1-2x} \\ &= \lim_{x\to 0}\frac{-2}{1-2x}\\ &= -2. \end{align*} Hence $\begin{align*} \lim_{x\to 0}(1-2x)^{1/x} &= \lim_{x\to 0} e^{\ln(1-2x)/x}\\ &= e^{\lim\limits_{x\to 0}\ln(1-2x)/x}\\ &= e^{-2}. \end{align*}$