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Vypo�t�te limitu:

$$\lim\limits_{x \to {\pi \over 4}} (\tan x)^{\tan(2x)}$$

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V�po�et dan� limity provedeme podle vzorce: $\lim\limits_{x\to a} u(x)^{v(x)} = e^{\lim\limits_{x\to a} [u(x)-1]v(x)}$. Plat�, �e $\lim\limits_{x \to {\pi \over 4}} \tan x = 1$.

$$\lim\limits_{x \to {\pi \over 4}} (\tan x -1)\tan(2x) = \lim\limits_{x \to {\pi \over 4}} \left( {\sin x \over \cos x} - 1 \right){\sin (2x) \over \cos (2x)}=$$

$$= \lim\limits_{x \to {\pi \over 4}} {\sin x  - \cos x \over \cos x} {2\sin x \cos x \over (\cos x - \sin x)(\cos x + \sin x )} = \lim\limits_{x \to {\pi \over 4}} {-2\sin x \over \cos x + \sin x} = -1$$

Hledan� limita je pak:

$$\lim\limits_{x \to {\pi \over 4}} (\tan x)^{\tan(2x)}={1 \over e}$$

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