Evaluate $\lim\limits_{x \to \frac{\pi}{4}} \; \left(\tan x\right)^{\tan 2x}$
Let $f\left(x\right)=\left(\tan x\right)^{\tan 2x}$
Then, $\log \left[f\left(x\right)\right] = \tan 2x \log \left(\tan x\right)$
$\begin{aligned}
\therefore \; \lim\limits_{x \to \frac{\pi}{4}} \; \left\{\log \left[f\left(x\right)\right] \right\} & = \lim\limits_{x \to \frac{\pi}{4}} \; \left[\tan 2x \log \left(\tan x\right)\right] & \left(\infty - 0 \text{ form}\right) \\
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& = \lim\limits_{x \to \frac{\pi}{4}} \left[\dfrac{\log \left(\tan x\right)}{\cot 2x}\right] & \left(\dfrac{0}{0} \text{ form}\right) \\
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& = \lim\limits_{x \to \frac{\pi}{4}}\; \dfrac{\dfrac{d}{dx}\left[\log \left(\tan x\right)\right]}{\dfrac{d}{dx}\left(\cot 2x\right)} & \left[\text{L'Hospital's rule}\right] \\
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& = \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{\sec^2 x / \tan x}{-2 \text{ cosec}^2 \; 2x} & \\
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& = \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{\dfrac{1}{\cos^2 x} \div \dfrac{\sin x}{\cos x}}{\dfrac{-2}{\sin^2 2x}} & \\
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& = \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{\sin^2 2x}{-2 \sin x \cos x} & \\
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& = - \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{\sin^2 2x}{\sin 2x} & \left[\text{Note: }\sin 2x = 2 \sin x \cos x\right] \\
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& = - \lim\limits_{x \to \frac{\pi}{4}} \; \sin 2x & \\
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& = - \sin \left(2 \times \dfrac{\pi}{4}\right) & \\
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& = - \sin \dfrac{\pi}{2} = -1
\end{aligned}$
$\therefore$ $\;$ $\lim\limits_{x \to \frac{\pi}{4}} \; \left\{\log \left[f\left(x\right)\right] \right\} = -1$
i.e. $\log \left\{\lim\limits_{x \to \frac{\pi}{4}} \left[f\left(x\right)\right] \right\} = -1$
i.e. $\lim\limits_{x \to \frac{\pi}{4}} \left(\tan x\right)^{\tan 2x} = e^{-1} = \dfrac{1}{e}$