3
$\begingroup$

I don't find the right identities for this

$$\lim _{x\to \frac{\pi }{4}}\left(\tan\left(2x\right)\tan\left(\frac{\pi }{4}-x\right)\right)$$

Someone can help me ? Thanks.

$\endgroup$
1
  • $\begingroup$ I have tried this, without success. $\endgroup$
    – NM2
    Jan 25, 2016 at 20:28

6 Answers 6

2
$\begingroup$

Since

\begin{align} \tan(2x)&=\frac{2\tan x}{1-\tan^2 x}&&&\text{and}&&\tan\left(\frac{\pi}{4}-x\right)&=\frac{\tan\frac{\pi}{4}-\tan x}{1+\tan\frac{\pi}{4}\tan x}=\frac{1-\tan x}{1+\tan x} \end{align} we have \begin{align} \lim_{x\to\frac{\pi}{4}}\left(\tan\left(2x\right)\tan\left(\frac{\pi }{4}-x\right)\right)&=\lim_{x\to\frac{\pi}{4}}\frac{(2\tan x)(1-\tan x)}{(1-\tan x)(1+\tan x)^2}\\ &=\lim_{x\to\frac{\pi}{4}}\frac{2\tan x}{(1+\tan x)^2}\\ &=\frac{2}{(1+1)^2}\\ &=\boxed{\color{blue}{\frac{1}{2}}} \end{align}

$\endgroup$
2
$\begingroup$

For $x\neq \frac{\pi}{4}+k\pi/2$ and $x\neq \pi/2+k\pi$ ($k\in\mathbb{Z}$), we have

$$\tan(2x)=\frac{2\tan(x)}{1-\tan^{2}(x)}$$

and

$$\tan\left(\frac{\pi}{4}-x\right)=\frac{\tan\left(\frac{\pi}{4}\right)-\tan(x)}{1+\tan\left(\frac{\pi}{4}\right)\tan(x)}$$

Combining the two expressions and using the fact that $\tan(\pi/4)=1$, we get

\begin{align*} \tan(2x)\tan\left(\frac{\pi}{4}-x\right) &=\frac{2\tan(x)}{1-\tan^{2}(x)}\cdot\frac{1-\tan(x)}{1+\tan(x)}\\ &=\frac{2\tan(x)}{(1+\tan(x))^{2}}\\ \end{align*}

Hence,

$$\lim_{x\to\frac{\pi}{4}}\tan(2x)\tan\left(\frac{\pi}{4}-x\right)=\lim_{x\to\frac{\pi}{4}}\frac{2\tan(x)}{(1+\tan(x))^{2}}=\frac{2}{(1+1)^{2}}=\frac{1}{2}$$

$\endgroup$
1
$\begingroup$

Notice, let $\frac{\pi}{4}-x=t\implies t\to 0$ as $x\to \pi/4$ $$\lim_{x\to \pi/4}\tan 2x\tan\left(\frac{\pi}{4}-x\right)$$ $$=\lim_{t\to 0}\tan\left(\frac{\pi}{2}-2t\right)\tan (t)$$ $$=\lim_{t\to 0}\cot\left(2t\right)\tan (t)$$ $$=\lim_{t\to 0}\frac{\cos\left(2t\right)}{\sin (2t)}\tan (t)$$ $$=\frac 12\lim_{t\to 0}\frac{\frac{\tan (t)}{t}}{\frac{\sin (2t)}{2t}}\cdot \cos\left(2t\right)$$ $$=\frac 12\frac{\lim_{t\to 0}\left(\frac{\tan (t)}{t}\right)}{\lim_{t\to 0}\left(\frac{\sin (2t)}{2t}\right)}\cdot \lim_{t\to 0}\cos\left(2t\right)$$ $$=\frac 12\cdot \frac{1}{1}\cdot 1=\color{red}{\frac 12}$$

$\endgroup$
1
$\begingroup$

Take $y=x-\frac{\pi}{4}$ You have $$f(x)=\tan\left(2x\right)\tan\left(\frac{\pi }{4}-x\right)=-\frac{\sin(2y+\frac{\pi}{2})}{\cos(2y+\frac{\pi}{2})}\frac{\sin y}{\cos y}=\frac{\cos 2y}{\sin 2y}\frac{\sin y}{\cos y}=\frac{\cos 2y}{2\cos^2y}$$ Hence $$\lim\limits_{x \to \frac{\pi}{4}} f(x)=\frac{1}{2}$$

$\endgroup$
0
0
$\begingroup$

In THIS ANSWER, I showed that the tangent function satisfies the inequalities

$$\left|z\right| \le \left|\tan(z)\right|\le \left|\frac{z}{\cos(z)}\right| \tag 1$$

for $|z|<\pi/2$.

For the problem of interest, we note first that

$$\tan(2x)\tan(\pi/4-x)=\frac{\tan(x-\pi/4)}{\tan(2(x-\pi/4))}.$$

Then, using the inequalities in $(1)$, we obtain the upper bound for $0<x-\pi/4<\pi/4$

$$\begin{align} |\tan(2x)\tan(\pi/4-x)|&=\tan(2x)\tan(\pi/4-x)\\\\ &\le \left|\frac{\frac{(x-\pi/4)}{\cos(x-\pi/4)}}{2(x-\pi/4)}\right|\\\\ &=\frac{1}{2\cos(x-\pi/4)} \tag 2 \end{align}$$

and the lower bound

$$\begin{align} |\tan(2x)\tan(\pi/4-x)|&=\tan(2x)\tan(\pi/4-x)\\\\ &\ge \left|\frac{x-\pi/4}{{\frac{2(x-\pi/4)}{\cos(2(x-\pi/4))}}}\right|\\\\ &=\frac{\cos(x-\pi/4)}{2} \tag 3 \end{align}$$

Putting together $(2)$ and $(3)$ we obtain

$$ \frac{\cos(x-\pi/4)}{2} \le \tan(2x)\tan(\pi/4-x)\le \frac{1}{2\cos(x-\pi/4)}$$

whereupon applying the Squeeze Theorem results in the limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \pi/4}\tan(2x)\tan(\pi/4-x)=\frac12}$$

$\endgroup$
0
-1
$\begingroup$

hint: $\tan(2x) = \dfrac{\sin(2x)}{\cos(2x)}$ and write $P = \sin(2x)\cdot \dfrac{\tan\left(\dfrac{\pi}{4} - x\right)}{\cos(2x)}$ and use L'hospitale rule for the second factor.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .