# How can we calculate the following limit?

I came across this problem and spent a lot of time on this but couldn't figure it out.

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

Using a general approach for $$1^\infty$$ type, I could come to the following expression:

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

But I can't get further from this.

I was able to observe that this is a ($$\infty*0$$)form so maybe we could transform the whole parenthesis with the $$\arctan$$ in the denominator so that we could get a $$\frac{\infty}{\infty}$$ form which could enable us to use L'Hospital's rule here but I that didn't yield a good result.

Is there any better way to do this?

• Missing a "$ln$" in your second display. May 13, 2021 at 22:15
• @EricTowers My teacher didn't mention that to me so I'm not aware of that... I am only aware of that identity just like here: math.stackexchange.com/q/1987215/843043 May 14, 2021 at 9:30
• $f^g = \mathrm{e}^{g \ln f}$. May 14, 2021 at 22:21

You were going the right way. All that is left is to write the limit with $$\frac 1x$$ in the denominator, then one application of L'Hospital's rule.
Note that\begin{align}\lim_{x\to\infty}x\left(\frac{2\arctan x}\pi-1\right)&=\lim_{x\to\infty}\frac{\frac{2\arctan x}\pi-1}{\frac1x}\\&=\lim_{x\to\infty}\frac{\frac2{\pi(1+x^2)}}{-\frac1{x^2}}\\&=-\frac2\pi.\end{align}
$$y=\Big[\frac{2 }{\pi }\tan ^{-1}(x)\Big]^x \implies \log(y)=x \log\Big[\frac{2 }{\pi }\tan ^{-1}(x)\Big]$$ $$\tan ^{-1}(x)=\frac{\pi }{2}-\frac{1}{x}+\frac{1}{3 x^3}+O\left(\frac{1}{x^5}\right)$$ $$\frac{2 }{\pi }\tan ^{-1}(x)=1-\frac{2}{\pi x}+\frac{2}{3 \pi x^3}+O\left(\frac{1}{x^5}\right)$$ $$\log\Big[\frac{2 }{\pi }\tan ^{-1}(x)\Big]=-\frac{2}{\pi x}-\frac{2}{\pi ^2 x^2}+O\left(\frac{1}{x^3}\right)$$ $$\log(y)=-\frac{2}{\pi }-\frac{2}{\pi ^2 x}+O\left(\frac{1}{x^2}\right)$$ $$y=e^{\log(y)}=e^{-2/\pi }\Big[1- \frac{2}{\pi ^2 x}\Big]+O\left(\frac{1}{x^2}\right)$$