How to prove that $\lim_{x\to \pi/2} (\tan x)^{\tan2x}=1$

Question

How do I prove that the limit of the following function is $1$?

$\displaystyle\lim_{x\to \pi/2} (\tan x)^{\tan2x}$

I shouldn't use l'Hôpital's rule.

I have proven that limit is greater than $1$ but couldn't sandwich on the other side (I mean, the limit is less than $1$).

Answer 1

Let $y=log(tan x)$ $\lim_{x\to{\pi/2}}$

Then $\lim_{x\to\pi/2}log((tan (x))^{tan2x})=\lim_{x\to\pi/2}tan2x(\log {tan (x)})$

Using $tan2x=2 tan (x)/(1+tan^2x))$

$\lim_{x\to\pi/2}tan2x(\log {tan (x)})=\lim_{y\to\infty}\frac{2e^y}{1-e^{2y}}y$

$=\frac{2y}{e^{-y}-e^y}=\frac{2}{\frac{e^{-y}}{y}-\frac{e^y}{y}}$

But

$\lim_{y\to\infty}\frac{e^{-y}}{y}=0$

$\lim_{y\to\infty}\frac{e^{y}}{y}=\infty$ as $e^y/y=1/y+1+y/2!...$ by Taylor expansion

So

$\lim_{y\to\infty}\frac{2}{\frac{e^{-y}}{y}-\frac{e^y}{y}}=0$ $\implies \lim_{x\to\pi/2}log((tan (x))^{tan2x})=0$ $\implies \lim_{x \to \pi/2}tan(x)^{tan 2x}=1$

Answer 2

Since $\tan x<0$ for $\pi/2<x<\pi$ we can assume the function is defined only for $0<x<\pi/2$ (a left neighborhood of $\pi/2$).

In all these case it's better to compute the limit of the logarithm: $$ \lim_{x\to\frac{\pi}{2}}\tan2x\log\tan x= \lim_{x\to\frac{\pi}{2}}-\frac{\log\cot x}{\cot2x}= \lim_{x\to\frac{\pi}{2}}-\frac{2\cot x\log\cot x}{\cot^2x-1} $$ by applying $\tan x=\frac{1}{\cot x}$ and the duplication formula $$ \cot2x=\frac{\cot^2x-1}{2\cot x}. $$ Now you can substitute $t=\cot x$ and get $$ \lim_{t\to0}\frac{2t\log t}{1-t^2} $$ and you should know how to manage this form. Hint: the limit is $0$.