Answer
$$\lim_{x\to0}(1-2x)^{1/x}=\frac{1}{e^2}$$
Work Step by Step
$$A=\lim_{x\to0}(1-2x)^{1/x}$$
We take the natural logarithm of both sides to take down the power of $\frac{1}{x}$.
$$\ln A=\ln[\lim_{x\to0}(1-2x)^{1/x}]$$
$$\ln A=\lim_{x\to0}[\ln(1-2x)^{1/x}]$$
$$\ln A=\lim_{x\to0}\frac{1}{x}\ln(1-2x)$$
$$\ln A=\lim_{x\to0}\frac{\ln(1-2x)}{x}$$
As $x$ approaches $0$, $\ln(1-2x)\to\ln(1-2\times0)=\ln1=0$.
That means we have an indeterminate form of $\frac{0}{0}$, and we use L'Hospital's Rules here:
$$\ln A=\lim_{x\to0}\frac{[\ln(1-2x)]'}{x'}$$
$$\ln A=\lim_{x\to0}\frac{\frac{1}{1-2x}(1-2x)'}{1}$$
$$\ln A=\lim_{x\to0}\frac{1}{1-2x}\times(-2)$$
$$\ln A=\lim_{x\to0}\frac{-2}{1-2x}$$
$$\ln A=\frac{-2}{1-2\times0}$$
$$\ln A=\frac{-2}{1}=-2$$
$$A=e^{-2}=\frac{1}{e^2}$$
