Найти предел: \(
\lim_{x\to1} x^{\frac{1}{1-x}}.
\)

Решение

\(\displaylines{
\lim_{x\to1} x^{\frac{1}{1-x}}
= \lim_{x\to1} e^{\ln x^{\frac{1}{1-x}}}
= \lim_{x\to1} e^{\frac{1}{1-x}\ln x } = \\
= \lim_{x\to1} e^{\frac{\ln x}{1-x}}
= e^{\lim_{x\to1} \frac{\ln x}{1-x} }
= e^{\lim_{x\to1} \frac{\frac{1}{x}}{-1} } = \\
= e^{\lim_{x\to1} \frac{1}{-x} }
= e^{-1} = \frac{1}{e} }
\)

Ответ: \( \large \frac{1}{e} \).