Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The limit is: $$\lim_{x\to\infty}\left(\frac{2\arctan(x)}{\pi}\right)^x$$ The limit is the kind of $1^\infty$, since: $$\lim_{x\to\infty}\left(\frac{2\arctan(x)}{\pi}\right)^x=\left(\frac{2\times\frac{\pi}{2}}{\pi}\right)^\infty=\left(\frac{\pi}{\pi}\right)^\infty=1^\infty$$ This is the farthest i went: $$L = \lim_{x\to\infty}\left(\frac{2\arctan(x)}{\pi}\right)^x$$ $$\ln(L) = \lim_{x\to\infty}\ln\left(\frac{2\arctan(x)}{\pi}\right)^x=\lim_{x\to\infty}x·\ln\left(\frac{2\arctan(x)}{\pi}\right)$$ Which becomes $0·\infty$ and I dont know where to go from here

share|improve this question

5 Answers 5

up vote 3 down vote accepted

There is a way to complete your argument, although the work is a little tedious. Observe that $$ x \cdot \ln \left( \frac{2 \cdot {\tan^{-1}}(x)}{\pi} \right) = \frac{\ln \left( \dfrac{2 \cdot {\tan^{-1}}(x)}{\pi} \right)}{\left( \dfrac{1}{x} \right)}. $$ As $$ \lim_{x \to \infty} \ln \left( \frac{2 \cdot {\tan^{-1}}(x)}{\pi} \right) = \ln \left( \lim_{x \to \infty} \frac{2 \cdot {\tan^{-1}}(x)}{\pi} \right) = \ln(1) = 0 $$ and $$ \lim_{x \to \infty} \frac{1}{x} = 0, $$ we can apply l’Hôpital’s Rule to obtain \begin{align} \ln(L) &= \lim_{x \to \infty} x \cdot \ln \left( \frac{2 \cdot {\tan^{-1}}(x)}{\pi} \right) \\ &= \lim_{x \to \infty} \frac{\ln \left( \dfrac{2 \cdot {\tan^{-1}}(x)}{\pi} \right)}{\left( \dfrac{1}{x} \right)} \\ &= \lim_{x \to \infty} \frac{\dfrac{d}{dx} \left[ \ln \left( \dfrac{2 \cdot {\tan^{-1}}(x)}{\pi} \right) \right]}{\dfrac{d}{dx} \left[ \dfrac{1}{x} \right]} \\ &= \lim_{x \to \infty} \frac{\left[ \dfrac{1}{(1 + x^{2}) \cdot {\tan^{-1}}(x)} \right]}{\left( - \dfrac{1}{x^{2}} \right)} \\ &= - \frac{2}{\pi}. \end{align} Therefore, $$ L = e^{\ln(L)} = e^{- 2/\pi}. $$

share|improve this answer
    
Thanks everyone for the answers, I've upvoted every one. Would be more correct to say $L=e^{\frac{-2}{/\pi}+2\pi k}$ ? –  Alejandro Jan 13 '13 at 13:54
    
@Alejandro: No, actually. The exponential function is injective, so changing the input exponent changes the output. –  Haskell Curry Jan 13 '13 at 20:12

Hint Make change of variables $$ 1+t=\frac{2\arctan(x)}{\pi} $$ and recall that $$ \cot\alpha\sim\frac{1}{\alpha}\quad\mbox{for}\quad\alpha\to 0 $$

share|improve this answer

Series expansion of $\arctan(x)$ about $\infty$ is $$\arctan(x) = \dfrac{\pi}2 - \dfrac1x + \mathcal{O} \left(\dfrac1{x^3}\right)$$ Hence, $$\lim_{x \to \infty}\left(\dfrac2{\pi} \arctan(x)\right)^x = \lim_{x \to \infty} \left(1- \dfrac2{\pi}\dfrac1x + \mathcal{O} \left(\dfrac1{x^3}\right)\right)^x = e^{-2/\pi}$$

share|improve this answer

Put $$\cot^{-1}x=y\implies \tan^{-1}x=\frac\pi2-y,y=\cot x$$ and as $x\to\infty,y\to 0$

$$\lim_{x\to\infty}\left(\frac{2\arctan(x)}{\pi}\right)^x$$

$$=\lim_{y\to 0}\left(\frac{2(\frac\pi2-y)}{\pi}\right)^{\cot y}$$

Let $z=\left(\frac{2(\frac\pi2-y)}{\pi}\right)^{\cot y}$

So, $$\log z=\cot y\log(1-\frac{2y}{\pi})$$

So, $$\lim_{y\to 0}\log z=\lim_{y\to 0}\cot y\log(1-\frac{2y}{\pi})=-\frac2\pi\frac{\left(\lim_{y\to 0}\cos y\right)}{\left(\lim_{y\to 0}\frac{\sin y}y\right)}\left(\lim_{y\to 0}\frac{\log(1-\frac{2y}{\pi})}{(-\frac{2y}{\pi})}\right)$$

$$\lim_{y\to 0}\log z=-\frac2\pi\implies \lim_{y\to 0}z=e^{-\frac2\pi}$$

share|improve this answer

Let's see a basic approach based upon $\lim_{x\to \infty} (1+1/x)^x=e$ and l'Hôpital's rule. Then

$$\lim_{x\to\infty}e^{\left(\displaystyle \frac{\arctan(x)-\frac{\pi}{2}}{\displaystyle\frac{\pi}{2x} }\right)}=\lim_{x\to\infty}e^{\left(\displaystyle \frac{-2x^2}{\pi (x^2+1)}\right)}=e^{\displaystyle\frac{-2}{\pi}}$$

Q.E.D. (Chris)

share|improve this answer
1  
Yay for Chris! ${}$ –  Parth Kohli Jan 13 '13 at 14:56
    
@DumbCow: heya :-) –  Chris's sis Jan 13 '13 at 14:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.