Calculate $\lim \limits_{x \to 0} (\tan ({\pi \over 4}-x))^{\cot x}$ without L'Hospital's Rule or Taylor expansion.

Question

Calculate $\lim \limits_{x \to 0} (\tan ({\pi \over 4}-x))^{\cot x}$

I tried to find some esoteric identities for $\tan ({\pi \over 4}-x)$, but I'd be glad if someone posted an elementary proof with basic trigonometric identities.

Answer 1

Hint: you can write the limit in this way: $\lim \limits_{x \to 0} e^{\cot x\cdot ln(\tan ({\pi \over 4}-x))}$ and then you can use this expression $\tan (u+v)= \frac {\tan u+ \tan v}{1-\tan u\cdot \tan v}$

Answer 2

If $L$ is the desired limit then \begin{align} \log L &= \log\left(\lim_{x \to 0}\tan\left(\frac{\pi}{4} - x\right)^{\cot x}\right)\notag\\ &= \lim_{x \to 0}\log\tan\left(\frac{\pi}{4} - x\right)^{\cot x}\text{ (by continuity of log)}\notag\\ &= \lim_{x \to 0}\cot x\log\tan\left(\frac{\pi}{4} - x\right)\notag\\ &= \lim_{x \to 0}\cot x\log\left(\frac{1 - \tan x}{1 + \tan x}\right)\notag\\ &= \lim_{x \to 0}\cot x\log\left(1 + \left(\frac{1 - \tan x}{1 + \tan x} - 1\right)\right)\notag\\ &= \lim_{x \to 0}\cot x\log\left(1 - \frac{2\tan x}{1 + \tan x}\right)\notag\\ &= \lim_{x \to 0}\cot x\dfrac{\log\left(1 - \dfrac{2\tan x}{1 + \tan x}\right)}{-\dfrac{2\tan x}{1 + \tan x}}\cdot\left(-\dfrac{2\tan x}{1 + \tan x}\right)\notag\\ &= -2\lim_{x \to 0}\frac{1}{1 + \tan x}\lim_{z \to 0}\frac{\log(1 + z)}{z}\text{ (putting }z = (-2\tan x)/(1 + \tan x))\notag\\ &= -2\cdot\frac{1}{1 + 0}\cdot 1 = -2 \end{align}

Thus desired limit $L = 1/e^{2}$.