Vtyjkyuk

$$lim_xto pi/4(tan x)^tan 2x $$

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

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

$$lim _ xto pi /4 (tan x)^ tan 2x = e ^ -1 $$

Take Log we have: $tan(2x)ln(tan x)= dfrac2tan xln(tan x)1-tan^2 x$. Put $u = tan x$, then the limit $l = 2displaystyle lim_u to 1dfracln u1-u^2=-1$, by L'hospitale rule. Thus the original limit $L = e^l = e^-1 = dfrac1e$

Set $x=pi/4-t$ so the base becomes
$$
tanleft(fracpi4-tright)=frac1-tan t1+tan t
$$
and the exponent is
$$
tan2x=cot2t=frac1-tan^2t2tan t
$$
so we really have to compute
$$
lim_tto0left(frac1-tan t1+tan tright)^1/tan t
$$
and then take its square root.

Let's split it into the limit from the right and from the left. For $tto0^+$, let's set $tan t=1/u$, so the limit becomes
$$
lim_utoinftyleft(fracu-1u+1right)^u=
lim_utoinftyleft(1-frac2u+1right)^u=
lim_utoinftyleft(1-frac2u+1right)^u+1
left(1-frac2u+1right)^-1
$$
Can you finish and also do the limit from the left?