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Question Number 83919 by john santu last updated on 08/Mar/20

$$\underset{x\to +\mathrm{\infty }}{\lim }{\left(\frac{3}{\pi }\mathrm{arc}\tan x\right)}^{2x}=?$$

Commented by john santu last updated on 08/Mar/20

$$=\mathrm{e}{}^{\underset{x\to +\mathrm{\infty }}{\lim }\left(\ln {\left(\frac{3}{\pi }{\tan }^{-1}\left(x\right)\right)}^{2x}\right)}$$$$={\mathrm{e}}^{\underset{x\to +\mathrm{\infty }}{\lim }\left(\frac{\ln \left(\frac{3}{\pi }{\tan }^{-1}\left(x\right)\right)}{\frac{1}{2x}}\right)}$$$$={\mathrm{e}}^{\underset{x\to +\mathrm{\infty }}{\lim }\frac{\pi }{{3\tan }^{-1}\left(x\right)}.\frac{3}{\pi (1+x^2)}.(-2x^2)}$$$$={\mathrm{e}}^{\underset{x\to +\mathrm{\infty }}{\lim }\frac{\pi }{3\left(\frac{\pi }{2}\right)}\times \underset{x\to +\mathrm{\infty }}{\lim }\frac{-6x^2}{\pi +\pi x^2}}$$$$={\mathrm{e}}^{\frac{2}{3}\times (-\frac{6}{\pi })}={\mathrm{e}}^{-\frac{4}{\pi }}$$$$=\frac{1}{\sqrt[\pi ]{{\mathrm{e}}^4}}$$

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