Question

Без Лопиталя как решать? ​

Answer 1

$\displaystyle \lim_{x \to \tfrac{\pi}{4} } (\text{tg} \, x)^{\text{tg} \, 2x} = [1^{\infty}] = \lim_{x \to \tfrac{\pi}{4} } (\text{tg} \, x)^{\tfrac{2 \, \text{tg} \, x}{1 - \text{tg}^{2} \, x} } = \left | {{\text{tg} \, x = 1 + t} \atop {t \to 0}} \right| =$

$\displaystyle = \lim_{t \to 0} (1 + t)^{\tfrac{2(1 + t)}{1 - (1 + t)^{2}} } =\lim_{t \to 0} (1 + t)^{-\tfrac{2 + 2t}{2t + t^{2}} } = \lim_{t \to 0} \left((1 + t)^{\tfrac{1}{t}}\right)^{ -\tfrac{2 + 2t}{2 + t} }} =$

$= e^{\displaystyle \lim_{t \to 0}\left(-\frac{2 + 2t}{2 + t} \right)} = e^{-1} = \dfrac{1}{e}$

Ответ: $\dfrac{1}{e}$