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

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

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