Задача Кузнецов Пределы 19-3

Условие задачи

Вычислить предел функции:

$\lim_{x\to \frac{\pi}{4}} \left( \frac{\ln {\left(\operatorname{tg}{x}\right)}}{1-\operatorname{ctg}{x}} \right)^{1 / \left(x+\frac{\pi}{4}\right)}$

Решение

$\lim_{x\to \frac{\pi}{4}} \left( \frac{\ln {\left(\operatorname{tg}{x}\right)}}{1-\operatorname{ctg}{x}} \right)^{1 / \left(x+\frac{\pi}{4}\right)} = \left(\lim_{x\to \frac{\pi}{4}} \frac{\ln {\left(\operatorname{tg}{x}\right)}}{1-\operatorname{ctg}{x}} \right)^{\lim\limits_{x\to \frac{\pi}{4}} 1 / \left(x+\frac{\pi}{4}\right)} =$

$= \left(\lim_{x\to \frac{\pi}{4}} \frac{\ln {\left(\operatorname{tg}{x}\right)}}{1-\operatorname{ctg}{x}} \right)^{1 / \left(\frac{\pi}{4}+\frac{\pi}{4}\right)} = \left(\lim_{x\to \frac{\pi}{4}} \frac{\ln {\left(\operatorname{tg}{x}\right)}}{1-\operatorname{ctg}{x}} \right)^{1 / \left(\frac{\pi}{2}\right)} =$

$= \left(\lim_{x\to \frac{\pi}{4}} \frac{\ln {\left(\operatorname{tg}{x}\right)}}{1-\operatorname{ctg}{x}} \right)^{\frac{2}{\pi}} = \left(\lim_{x\to \frac{\pi}{4}} \frac{\ln {\left(1-\left(1-\operatorname{tg}{x}\right)\right)}}{1-\operatorname{ctg}{x}} \right)^{\frac{2}{\pi}} =$

Воспользуемся заменой эквивалентных бесконечно малых:

$\ln {\left(1-\left(1-\operatorname{tg}{x}\right)\right)} \sim -\left(1-\operatorname{tg}{x}\right)$, при $x \to \frac{\pi}{4}\left(-\left(1-\operatorname{tg}{x}\right)\to 0\right)$

Получаем:

$= \left(\lim_{x\to \frac{\pi}{4}} \frac{-\left(1-\operatorname{tg}{x}\right)}{1-\operatorname{ctg}{x}} \right)^{\frac{2}{\pi}} = \left(\lim_{x\to \frac{\pi}{4}} \frac{\operatorname{tg}{x}-1}{1-\operatorname{ctg}{x}} \right)^{\frac{2}{\pi}} =$

$= \left(\lim_{x\to \frac{\pi}{4}} \frac{\frac{1}{\operatorname{ctg}{x}}-\frac{\operatorname{ctg}{x}}{\operatorname{ctg}{x}}}{1-\operatorname{ctg}{x}} \right)^{\frac{2}{\pi}}= \left(\lim_{x\to \frac{\pi}{4}} \frac{\frac{1}{\operatorname{ctg}{x}}\left(1-\operatorname{ctg}{x}\right)}{1-\operatorname{ctg}{x}} \right)^{\frac{2}{\pi}} =$

$= \left(\lim_{x\to \frac{\pi}{4}} \operatorname{tg}{x} \right)^{\frac{2}{\pi}} = \left( \operatorname{tg}{\frac{\pi}{4}} \right)^{\frac{2}{\pi}} = 1^{\frac{2}{\pi}} =1$