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