H ow to solve this limit ? $\lim_{x\to \pi/4}{(\tan x)^{\tan 2x}} $

Question

$$\lim_{x\to \pi/4}{(\tan x)^{\tan 2x}} $$

I think this is the second remarkable limit, but here is how to solve it - I do not know. help me please

Answer 1

$$\lim _{ x\to \pi /4 }{ (\tan x)^{ \tan 2x } } \\ x-\frac { \pi }{ 4 } =t\\ \lim _{ t\rightarrow 0 }{ { \left( \tan { \left( t+\frac { \pi }{ 4 } \right) } \right) }^{ \tan { 2\left( t+\frac { \pi }{ 4 } \right) } } } =...\\ \\ \tan { \left( t+\frac { \pi }{ 4 } \right) } =\frac { \tan { t } +1 }{ 1-\tan { t } } =1+\frac { 2\tan { t } }{ 1-\tan { t } } \\ \tan { \left( 2t+\frac { \pi }{ 2 } \right) } =-\cot { 2t } \\ \lim _{ t\rightarrow 0 }{ { \left( \tan { \left( t+\frac { \pi }{ 4 } \right) } \right) }^{ \tan { 2\left( t+\frac { \pi }{ 4 } \right) } } } =\lim _{ t\rightarrow 0 }{ { \left( 1+\frac { 2\tan { t } }{ 1-\tan { t } } \right) }^{ -\cot { 2t } } } =\\ =\lim _{ t\rightarrow 0 }{ { \left[ { { \left( 1+\frac { 2\tan { t } }{ 1-\tan { t } } \right) } }^{ \frac { 1-\tan { t } }{ 2\tan { t } } } \right] }^{ \frac { 2\tan { t } }{ 1-\tan { t } } \quad \cdot \left( -\cot { 2t } \right) } } ={ e }^{ \lim _{ t\rightarrow 0 }{ \frac { 2\tan { t } }{ 1-\tan { t } } \quad \cdot \left( -\cot { 2t } \right) } }=...\\ \lim _{ t\rightarrow 0 }{ \frac { 2\tan { t } }{ 1-\tan { t } } \quad \cdot \left( -\cot { 2t } \right) } =-\lim _{ t\rightarrow 0 }{ \frac { 2\sin { t } }{ \cos { t-\sin { t } } } \quad \cdot \frac { \left( \cos { t } -\sin { t } \right) \left( \cos { t } +\sin { t } \right) }{ 2\sin { t\cos { t } } } } =-1\\ { e }^{ \lim _{ t\rightarrow 0 }{ \frac { 2\tan { t } }{ 1-\tan { t } } \quad \cdot \left( -\cot { 2t } \right) } }={ \quad e }^{ -1 }\\ $$

$$\lim _{ x\to \pi /4 }{ (\tan x)^{ \tan 2x } } ={ e }^{ -1 }$$

Answer 2

Take Log we have: $\tan(2x)\ln(\tan x)= \dfrac{2\tan x\ln(\tan x)}{1-\tan^2 x}$. Put $u = \tan x$, then the limit $l = 2\displaystyle \lim_{u \to 1}\dfrac{\ln u}{1-u^2}=-1$, by L'hospitale rule. Thus the original limit $L = e^l = e^{-1} = \dfrac{1}{e}$

Answer 3

Set $x=\pi/4-t$ so the base becomes $$ \tan\left(\frac{\pi}{4}-t\right)=\frac{1-\tan t}{1+\tan t} $$ and the exponent is $$ \tan2x=\cot2t=\frac{1-\tan^2t}{2\tan t} $$ so we really have to compute $$ \lim_{t\to0}\left(\frac{1-\tan t}{1+\tan t}\right)^{1/\tan t} $$ and then take its square root.

Let's split it into the limit from the right and from the left. For $t\to0^+$, let's set $\tan t=1/u$, so the limit becomes $$ \lim_{u\to\infty}\left(\frac{u-1}{u+1}\right)^u= \lim_{u\to\infty}\left(1-\frac{2}{u+1}\right)^u= \lim_{u\to\infty}\left(1-\frac{2}{u+1}\right)^{u+1} \left(1-\frac{2}{u+1}\right)^{-1} $$ Can you finish and also do the limit from the left?