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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

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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, $$ L = e^{\ln(L)} = e^{- 2/\pi}. $$

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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 $$ \cot\alpha\sim\frac{1}{\alpha}\quad\mbox{for}\quad\alpha\to 0 $$

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

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


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

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

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

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

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

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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)

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Yay for Chris! ${}$ – Parth Kohli Jan 13 '13 at 14:56
@DumbCow: heya :-) – Chris's sis the artist Jan 13 '13 at 14:57

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