I`m trying to evaluate this limit and I need some advice how to do that.
$$\displaystyle\lim_{x\to \infty}\left(\frac{2\arctan(x)}{\pi}\right)^x $$
I have a feeling it has to do with a solution of form $1^\infty$ but do not know how to proceed. Any hints/solutions/links will be appreciated
Real Analysis – What is the Limit of (2arctan(x)/?)^x as x Approaches Infinity?
calculuslimitsreal-analysis
Best Answer
It could be suitably modified into something involving the limit $(1+\frac1x)^x\rightarrow e$ for $x\to\infty$. $$ \left(\frac{2\arctan x}{\pi}\right)^x ~=~ \left[1 + \left(\frac{2\arctan x}{\pi}-1\right)\right]^x $$ Let $f(x)=\left(\frac{2\arctan x}{\pi}-1\right)$; clearly $f(x)\to 0$ for $x\to+\infty$, therefore $$ \left[1+f(x)\right]^{\frac{1}{f(x)}}\longrightarrow e $$ Let us focus on the limit of $xf(x)$: using l'Hospital's rule we get $$ \lim_{x\to+\infty}x\,f(x) ~=~ \lim_{x\to+\infty} \frac{\frac{2\arctan x-1}{\pi}}{\frac1x} ~\stackrel H=~ \lim_{x\to+\infty} \frac{\frac{2}{\pi(1+x^2)}}{-\frac1{x^2}} ~=~ -\frac2\pi $$ Now, putting all together: $$ \lim_{x\to+\infty} \left(\frac{2\arctan x}{\pi}\right)^x ~=~ \lim_{x\to+\infty} \big(1+f(x)\big)^x ~=~ \lim_{x\to+\infty} \left[\big(1+f(x)\big)^{\frac{1}{f(x)}}\right]^{xf(x)} ~=~ e^{-2/\pi} $$ Generally, when you run into $1^\infty$ you can work it out in this way.