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I am not sure what should I use. series for $\arctan x$ or L'hopital's rules cause it converges to $\infty$

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

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    What is "arctgx"? Is this "arc tangent of x"? Or something else? – 2017-02-20
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    it might be interesting when $\lim _{ x\to \infty } (\frac { 2 }{ \pi } arctgx)^{ x }$ – 2017-02-20
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    If *it really is* $\;\frac\pi2\;$ inside the parentheses then the limit is almost trivial infinity, since then $\;\arctan x\to\frac\pi2\;$ , so for $\;x\;$ big enough we have inside the parentheses $\;\frac\pi2(\frac\pi2-0.001)>1\implies\;$ the limit is infinity. – 2017-02-20
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    @M.Mass Are you sure this is the problem? – 2017-02-20
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    Did you make typo? Is it$$\lim_{x\to \infty} \left( \color{red}{\frac{2}{\pi}} \tan^{-1} x \right)^{x} ?$$ – 2017-02-20

3 Answers 3

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$$ \lim_{x \to \infty} \left(\frac2\pi\arctan x\right)^x=\lim_{t\to 0^+} \left(\frac2\pi\text{arccot} t\right)^{1/t}=\lim_{t\to 0^+} \left(1-\frac2\pi\arctan t\right)^{1/t}\\ =\lim_{u\to 0^+} \left(1-\frac2\pi u\right)^{1/\tan u} =\lim_{u\to 0^+} \left(\left(1-\frac2\pi u\right)^{1/u}\right)^{u/\tan u}\\ =e^{-2/\pi}.$$

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    See @haqnatural's comment. – 2017-02-20
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When $x\to +\infty$ we have that $\arctan x \to \frac {\pi} 2$ thus $\lim_{x\to +\infty}\left(\frac {\pi^2} 4\right)^x=+\infty$ because $\frac {\pi^2} 4 \gt \frac {3^2} 4 \gt1$

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Assuming we are looking at $[(2/\pi)\arctan x]^x,$ apply $\ln$ to get

$$x\ln [(2/\pi)\arctan x] = \frac{\ln [(2/\pi)\arctan x]}{1/x}.$$

As $x\to \infty,$ the numerator and denominator both $\to 0,$ so let's try L'Hopital. The quotient of derivatives equals

$$\tag 1\frac{1}{(2/\pi)\arctan x}\cdot\frac{2/\pi}{1+x^2}\cdot\frac{1}{-1/x^2}.$$

The first fraction $\to 1,$ while the product of the last two fractions $\to -2/\pi.$ Thus the limit of $(1)$ is $-2/\pi.$ Exponentiating back then gives $e^{-2/\pi}$ for the original limit.