Найти предел функции (фото во вложении)

$\lim\limits _{x \to \frac{\pi}{4}}\, (tgx)^{tg2x}= \lim\limits _{x \to \frac{\pi }{4} }\, (tgx)^{\frac{2tgx}{1-tg^2x}}=[1+t=tgx\; ,\; t\to 0\; ,\; tg\frac{\pi}{4}=1\, ]=\\\\= \lim\limits _{t \to 0}\; (1+t)^{ \frac{2(1+t)}{1-(1+t)^2} }= \lim\limits _{t \to 0}\, (1+t)^{-\frac{2+2t}{-t^2-2t}}= \lim\limits _{t \to 0}(1+t)^{-\frac{2t+2}{t(t+2)}}=\\\\=\lim\limits _{t \to 0}\, \Big ((1+t)^{\frac{1}{t}}\Big )^{-\frac{2t+2}{t+2}}=e^{\lim\limits _{t \to 0}\, \frac{-2t-2}{t+2}}=[ \frac{-2\cdot 0-2}{0+2}=-1 ]=e^{-1}= \frac{1}{e}$

или

$\lim\limits _{x \to \frac{\pi}{4}}\, (tgx)^{tg2x}=\lim\limits _{x \to \frac{\pi}{4}}\Big (1+(tgx-1)\Big )^{\frac{2tgx}{1-tg^2x}}=\\\\=\lim\limits_{x \to \frac{\pi }{4}}\, \Big (\Big (1+(tgx-1)\Big )^{\frac{1}{tgx-1}}\Big )^{(tgx-1)\cdot \frac{2tgx}{1-tg^2x}}=\\\\=e^{\lim\limits _{x \to \frac{\pi}{4}}\,\frac{-(1-tgx)\cdot 2tgx}{(1-tgx)(1+tgx)}}=e^{\lim\limits _{x \to \frac{\pi}{4}}\, \frac{-2tgx}{tgx+1}}=e^{ \frac{-2\cdot 1}{1+1}}=e^{-1}= \frac{1}{e}$