Evaluation of x^(1/(1-x))

$lim_{x \rightarrow 1} x^{\frac{1}{1-x}}$
is of indeterminate form, so we can use l'Hospital's Rule to evaluate it.
Start by taking logs.
$ln(x^{\frac{1}{1-x}})=\frac{1}{1-x} lnx$

$lim_{x \rightarrow 1}\frac{1}{1-x} lnx$
is again of indeterminate form. L'Hospital's Rule gives
\begin{aligned} lim_{x \rightarrow 1} \frac{lnx}{1-x} &= lim_{x \rightarrow 1} \frac{1/x}{-1} \\ &= -1 \end{aligned}

Hence
$lim_{x \rightarrow 1} x^{\frac{1}{1-x}}=e^{-1}=\frac{1}{e}$