2022-04-01

I am trying to find this limit but I failed.
$\underset{x\to \mathrm{\infty }}{lim}\left(2\mathrm{arctan}x-\pi \right)\mathrm{ln}x$

Marcos Boyer

$\left(2\mathrm{arctan}\left(x\right)-\pi \right)\mathrm{ln}\left(x\right)=-2\left(\frac{\pi }{2}-\mathrm{arctan}\left(x\right)\right)\mathrm{ln}\left(x\right)=-2\mathrm{arctan}\left(\frac{1}{x}\right)\mathrm{ln}\left(x\right)$
$\underset{x\to \mathrm{\infty }}{lim}\left(2\mathrm{arctan}\left(x\right)-\pi \right)\mathrm{ln}\left(x\right)=-2\underset{x\to \mathrm{\infty }}{lim}\mathrm{arctan}\left(\frac{1}{x}\right)\mathrm{ln}\left(x\right)=2\underset{x\to {0}^{+}}{lim}\mathrm{arctan}\left(x\right)\mathrm{ln}\left(x\right)$
$\underset{x\to {0}^{+}}{lim}\mathrm{arctan}\left(x\right)\mathrm{ln}\left(x\right)=\underset{x\to {0}^{+}}{lim}\frac{\mathrm{arctan}\left(x\right)}{\frac{1}{\mathrm{ln}\left(x\right)}}=\underset{x\to {0}^{+}}{lim}\frac{\frac{1}{1+{x}^{2}}}{-\frac{1}{{\mathrm{ln}}^{2}\left(x\right)}\frac{1}{x}}=-\underset{x\to {0}^{+}}{lim}\frac{x{\mathrm{ln}}^{2}\left(x\right)}{1+{x}^{2}}=0$
Hence,
$\underset{x\to \mathrm{\infty }}{lim}\left(2\mathrm{arctan}\left(x\right)-\pi \right)\mathrm{ln}\left(x\right)=0$

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