# How to determine this limit without L'Hospital? [closed]

How to solve this limit without using L'Hospital?

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

## closed as off-topic by GNUSupporter 8964民主女神 地下教會, Shailesh, Saad, JMP, Cave JohnsonApr 25 '18 at 12:44

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – GNUSupporter 8964民主女神 地下教會, Shailesh, Saad, JMP, Cave Johnson
If this question can be reworded to fit the rules in the help center, please edit the question.

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

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