How to find lim $(\tan x)^{\tan(2x)}$, $x\to\pi/2.$

Question

The limit

Important condition: you can't use l'Hopital's rule.

I've been thinking about that for an hour, but still can't do it. I need a full solution. Wolfram says that the answer is $1.$

Thanks in advance.

Answer 1

Correct me if wrong:

Note:$ \tan(2x)=\dfrac{2\tan x}{1-tan^2 x}.$

Take $ \log $ of your expression:

$\tan 2x \log (\tan x)$

$f(y):= \dfrac{2y \log y}{1-y^2}$, where $y:=\tan x$.

$ \lim_{y \rightarrow \infty} f(y)= $

$\lim_{y \rightarrow \infty} 2\dfrac{y\log y}{y^2(1/y^2 -1)}=$

$\lim_{y \rightarrow \infty} 2\dfrac{\log y}{y(1/y^2-1)}=0.$

To get back to the original limit consider:

$ \lim_{y \rightarrow \infty} e^{f(y)}=$

$e^{ \lim_{y \rightarrow \infty} f(y)}=1.$

(Continuity of the exponential function).

Used $\lim_{y \rightarrow \infty} \dfrac{\log y}{y}=0$.

Answer 2

Let $y=\pi/2-x\to 0$ and recall that

$$\tan x=\frac1{\tan(\pi/2-x)}=\frac1{\tan y}$$

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

and then by $z=\tan y\to 0$ we have

$$\lim_{x\to \pi/2} (\tan x)^{\tan 2x}=\lim_{z\to 0} \left(\frac1z\right)^\frac{2z}{z^2-1}=\lim_{z\to 0} \left(\frac1{z^z}\right)^\frac{2}{z^2-1}=\left(\frac1{1}\right)^{-2}=1$$

Answer 3

Hint: if $f(x) = (\tan x)^{\tan (2x)}$, then $\ln f(x) = \tan(2x) \ln(\tan x) = \frac{\ln(\tan x)}{\cot(2x)}$. Now, you should be able to apply l'Hopital's rule to this quotient, then do some simplification to find $\lim_{x \to \pi/2} [\ln f(x)]$.

Answer 4

Calculate first the easier limit $$\lim_{x\to\frac{\pi}{2}}(1+\tan(x))^\frac{1}{\tan(x)}=1$$ Then do you have the expression $$(1+\tan(x))^{\tan(2x)}=(1+\tan(x))^{\frac{\tan(x)}{\tan(x)}\tan(2x)}=(1+\tan(x))^{\frac{1}{\tan(x)}\cdot\frac{2\tan^2(x)}{1-\tan^2(x)}}$$

Verify now that $$\lim_{x\to\frac{\pi}{2}}\frac{2\tan^2(x)}{1-\tan^2(x)}=-2$$ Your limit is equal to $1^{-2}=1$