Answer
$$
\lim _{x \rightarrow 0} (1-2 x)^{1 / x}=e^{-2}
$$
Work Step by Step
$$
\lim _{x \rightarrow 0} (1-2 x)^{1 / x}
$$
Let
$$
y= (1-2 x)^{1 / x}
$$
Then
$$
\ln y=\frac{1}{x} \ln (1-2 x)= \frac{ \ln (1-2 x)}{x}
$$
notice that as $ \quad x \rightarrow 0 \quad $ we have $ \quad \ln (1-2 x) \rightarrow 0 \quad $ and $ \quad x \rightarrow 0 $
so l’Hospital’s Rule gives
$$
\begin{aligned}
\lim _{x \rightarrow 0} \ln y &=\lim _{x \rightarrow 0} \frac{\ln (1-2 x)}{x} \quad\quad \rightarrow \frac{0}{0}\\
& \stackrel{\mathrm{H}}{=} \lim _{x \rightarrow 0} \frac{-2 /(1-2 x)}{1}\\
&=-2
\end{aligned}
$$
Hence, we have
$$
\begin{aligned}
\lim _{x \rightarrow 0}(1-2 x)^{1 / x} &=\lim _{x \rightarrow 0} e^{\ln y} \\
&=e^{-2}\\
\end{aligned}
$$
