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How to solve this limit without using L'Hospital?

$$\lim_{x\to\ +∞} [(π-2\arctan{x})\log_{10}{x}$$

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closed as off-topic by GNUSupporter 8964民主女神 地下教會, Shailesh, Saad, JonMark Perry, Cave Johnson Apr 25 '18 at 12:44

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    $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. $\endgroup$ – José Carlos Santos Apr 25 '18 at 9:59
  • $\begingroup$ Try to substitute $x$ wirh $\tan y$, $y\to (\pi/2)^-$ (or maybe easier tractavble, $\cot y$, $y\to 0^+$) $\endgroup$ – Hagen von Eitzen Apr 25 '18 at 10:02
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Hint:

Factor out $2$ and use the formula $$\arctan x+\arctan\frac1x=\frac\pi2\quad\text{if }x>0.$$ You obtain $$(π-2\arctan{x})\log_{10}{x}=2\arctan \frac1x\,\frac{\ln x}{\ln 10}\sim_\infty 2\frac1x\frac{\ln x}{\ln 10}=\frac 2{\ln 10}\frac{\ln x}x\to 0.$$

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With $\dfrac1x=y$ $$\lim_{x\to\infty} 2(\dfrac{\pi}{2}-\arctan{x})\dfrac{\ln x}{\ln10}=\dfrac{2}{\ln10}\lim_{x\to\infty} \arctan\frac1x\ln x=\dfrac{-2}{\ln10}\lim_{y\to0} \arctan y\ln y$$ then with sandwich theorem $\ln y<y$ and $\arctan y<\dfrac{\pi}{2}$ the limit is $0$.

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