Задача №2034
Условие
Найти предел \(\lim_{x\to{1}}\frac{x^x-x}{\ln{x}-x+1}\).
Решение
\[
\lim_{x\to{1}}\frac{x^x-x}{\ln{x}-x+1}
=\left[\frac{0}{0}\right]
=\lim_{x\to{1}}\frac{\left(x^x-x\right)'}{\left(\ln{x}-x+1\right)'}=\\
=\lim_{x\to{1}}\frac{x^x(\ln{x}+1)-1}{\frac{1}{x}-1}
=\left[\frac{0}{0}\right]
=\lim_{x\to{1}}\frac{\left(x^x(\ln{x}+1)-1\right)'}{\left(\frac{1}{x}-1\right)'}
=\lim_{x\to{1}}\frac{x^x(\ln{x}+1)^2+x^{x}\cdot\frac{1}{x}}{-\frac{1}{x^2}}
=-2.
\]
Ответ:
\(-2\)