Answered: "I don't find the right identities for thislimx→π4(tan⁡(2x)tan⁡(π4−x))"

Pasegeabe85xy

Answered question

2022-03-30

I don't find the right identities for this

lim_{x \to \frac{π}{4}} (\tan (2 x) \tan (\frac{π}{4} - x))

Answer & Explanation

Jazlyn Mitchell

Beginner2022-03-31Added 14 answers

Since
$\tan (2 x) = \frac{2 \tan x}{1 - \tan^{2} x}$
and
$\tan (\frac{π}{4} - x) = \frac{\tan \frac{π}{4} - \tan x}{1 + \tan \frac{π}{4} \tan x} = \frac{1 - \tan x}{1 + \tan x}$
we have
$lim_{x \to \frac{π}{4}} ((\tan (2 x) \tan (\frac{π}{4} - x)) = lim_{x \to \frac{π}{4}} \frac{(2 \tan x) (1 - \tan x)}{(1 - \tan x) {(1 + \tan x)}^{2}}$
$= lim_{x \to \frac{π}{4}} \frac{2 \tan x}{{(1 + \tan x)}^{2}}$
$= \frac{2}{{(1 + 1)}^{2}}$
$= \frac{1}{2}$

Cason Harmon

Beginner2022-04-01Added 8 answers

Notice, let

\frac{π}{4} - x = t \Rightarrow t \to 0

x \to \frac{π}{4}

lim_{x \to \frac{π}{4}} \tan 2 x \tan (\frac{π}{4} - x)

= lim_{t \to 0} \tan (\frac{π}{2} - 2 t) \tan (t)

= lim_{t \to 0} \cos (2 t) \tan (t)

= \frac{1}{2} lim_{t \to 0} \frac{\cos (2 t)}{\sin (2 t)} \tan (t)

= \frac{1}{2} lim_{t \to 0} \frac{\frac{\tan (t)}{t}}{\frac{\sin 2 t}{2 t}} \cdot \cos (2 t)

= \frac{1}{2} \cdot \frac{1}{1} \cdot 1 = \frac{1}{2}

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