Answer
$\lim\limits_{x \to 1^+}x^{1/(1-x)}= \frac{1}{e}$
Work Step by Step
$\lim\limits_{x \to 1^+}x^{1/(1-x)}=\lim\limits_{x \to 1^+}(e^{ln~x})^{1/(1-x)}=\lim\limits_{x \to 1^+}e^{\frac{1}{(1-x)}~ln~x}$
$\lim\limits_{x \to 1^+}\frac{ln~x}{1-x} = \lim\limits_{x \to 1^+}\frac{(1/x)}{-1} = -1$
Therefore:
$\lim\limits_{x \to 1^+}e^{\frac{1}{(1-x)}~ln~x} = e^{-1} = \frac{1}{e}$
