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Compute the limit of the given function

by RoRi

Evaluate the limit.

    \[ \lim_{x \to \frac{\pi}{4}} (\tan x)^{\tan (2x)}. \]

Using the continuity and definition of the exponential we have

    \begin{align*} \lim_{x \to \frac{\pi}{4}} (\tan x)^{\tan (2x)} &= \lim_{x \to \frac{\pi}{4}} e^{ \tan (2x) \cdot \log (\tan x)} \\[9pt] &= \exp \left( \lim_{x \to \frac{\pi}{4}} \tan (2x) \cdot \log (\tan x) \right) \\[9pt] &= \exp \left( \lim_{x \to \frac{\pi}{4}} \left( \frac{\sin (2x)}{\cos (2x)} \cdot (\log (\cos x) - \log (\sin x)) \right) \right) \\[9pt] &= \exp \left( \lim_{x \to \frac{\pi}{4}} \frac{\sin (2x) \cdot (\log (\cos x) - \log (\sin x))}{\cos (2x)} \right). \end{align*}

Next, we apply L’Hopital’s rule to this limit,

    \begin{align*} &= \exp \left( \lim_{x \to \frac{\pi}{4}} \frac{ 2 \cos (2x) (\log (\sin x) - \log (\cos x)) + \sin (2x) \left( \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x}\right)}{-2 \sin (2x)} \right) \\[9pt] &= \exp \left( \lim_{x \to \frac{\pi}{4}} \frac{ 2 \cos (2x) \cdot (\log(\sin x) - \log(\cos x))}{-2 \sin (2x)} + \lim_{x \to \frac{\pi}{4}} \frac{ \sin (2x) \left( \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x} \right)}{-2 \sin (2x)}\right) \\[9pt] &= \exp \left( 0 + \lim_{x \to \frac{\pi}{4}} \frac{\frac{\cos^2 x + \sin^2 x}{\sin x \cos x}}{-2} \right) \\[9pt] &= \exp \left( \lim_{x \to \frac{\pi}{4}} \frac{\cos^2 x + \sin^2 x}{-2 \sin x \cos x} \right) \\[9pt] &= \exp \left( \lim_{x \to \frac{\pi}{4}} \frac{-1}{\sin (2x)} \right) \\[9pt] &= e^{-1} \\[9pt] &= \frac{1}{e}. \end{align*}

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