nice-calculus-I-lim-x-pi-4-tan-x-tan-2x-

Question Number 140046 by mnjuly1970 last updated on 03/May/21

\dots \dots \dots . . n i c e \dots \dots . c a l c u l u s (I) \dots \dots . .

Θ := {l i m}_{x \to \frac{π}{4}} {(t a n (x))}^{t a n (2 x)} = ?

\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots .

Commented by mnjuly1970 last updated on 03/May/21

t h a n k s a l o t \dots

Answered by Dwaipayan Shikari last updated on 03/May/21

\lim_{z \to 0} t a n (\frac{π}{2} + 2 z) l o g (t a n (\frac{π}{4} + z)) = l o g (y)

- c o t (2 z) l o g (\frac{1 + t a n z}{1 - t a n z}) = l o g (y)

\sim - c o t (2 z) l o g (1 + z) + c o t (2 z) l o g (1 - z) = l o g (y)

\approx \frac{- 1}{2 z} z + \frac{- z}{2 z} = l o g (y) \Rightarrow l o g (y) = - 1 \Rightarrow y = \frac{1}{e}

Answered by mnjuly1970 last updated on 04/May/21

s o l u t i o n \dots . .

t a n (x) := y \Rightarrow {_{y \to 1}^{x \to \frac{π}{4}}

Θ := {l i m}_{y \to 1} {(1 - (1 - y))}^{\frac{2 y}{(1 - y) (1 + y)}}

: \overset{1 - y = t}{=} {l i m}_{t \to 0} {(1 - t)}^{\frac{2 (1 - t)}{t (2 - t)}} = {l i m}_{t \to 0} {{(1 - t)}^{\frac{2}{t}}}^{\frac{1 - t}{2 - t}}

:= {(e^{- 2})}^{\frac{1}{2}} = e^{- 1} = \frac{1}{e} \dots \dots ✓ ✓

\dots \dots \dots Θ := \frac{1}{e} \dots \dots . .