\[lim_{x \rightarrow 1} x^{\frac{1}{1-x}}\]
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\[\begin{equation} \begin{aligned} lim_{x \rightarrow 1} \frac{lnx}{1-x} &= lim_{x \rightarrow 1} \frac{1/x}{-1} \\ &= -1 \end{aligned} \end{equation}\]
Hence
\[lim_{x \rightarrow 1} x^{\frac{1}{1-x}}=e^{-1}=\frac{1}{e}\]