(C) $\lim _{x \rightarrow e }(\log x)^{\frac{1}{1-\log x}}$ $=\lim _{x \rightarrow e }\left\{1+\log _{ e } x-1\right\}^{\frac{1}{1-\log x}}$ If $\lim _{x \rightarrow a } f (x)=1$ and $\lim _{x \rightarrow a } g (x)=\infty$, then $\lim _{x \rightarrow a }[ f (x)]^{ g (x)}= e ^{\lim _{x \rightarrow a } g (x)[ f (x)-1]}$ $=e^{\lim _{x \rightarrow e}\left(\log _{e} x-1\right) \times \frac{1}{1-\log _{e} x}}$ $= e ^{-1}$
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