# Find $\lim\limits_{x\to \infty} (2\arctan x -\pi)\ln x$

I am trying to find this limit but I failed.

I tried to split it and find that the two limits exist and then find an answer but with no luck. I also tried changing but it didn't help.

$$\lim_{x\to \infty} (2\arctan x -\pi)\ln x$$

Any hints?

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Did you try multiplying by 1/ln(x) on top and bottom? – The Chaz 2.0 Dec 9 '11 at 22:27
L'Hospital's Rule seems to work fine. – André Nicolas Dec 9 '11 at 22:35
@AndréNicolas you mean $\frac{(2\arctan x -\pi)}{\frac{1}{\ln x}}$? – practic_pom Dec 9 '11 at 22:38
@practic_pom: Yes, it is an automatic thing to do. – André Nicolas Dec 10 '11 at 0:02

$$(2 \arctan(x) - \pi) \ln(x) = - 2 (\pi/2 - \arctan(x)) \ln(x) = -2 \arctan(1/x) \ln(x)$$ $$\lim_{x \rightarrow \infty} (2 \arctan(x) - \pi) \ln(x) = -2 \lim_{x \rightarrow \infty} \arctan(1/x) \ln(x) = 2 \lim_{x \rightarrow 0^+} \arctan(x) \ln(x)$$ $$\lim_{x \rightarrow 0^+} \arctan(x) \ln(x) = \lim_{x \rightarrow 0^+}\frac{\arctan(x)}{\frac1{ \ln(x)}} = \lim_{x \rightarrow 0^+} \frac{\frac{1}{1+x^2}}{-\frac1{\ln^2(x)} \frac1x} = - \lim_{x \rightarrow 0^+} \frac{x \ln^2(x)}{1+x^2} = 0$$ Hence, $$\lim_{x \rightarrow \infty} (2 \arctan(x) - \pi) \ln(x) = 0.$$

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Maple has this asymptotic information for arctan as $x \to \infty$...

$$\arctan x \sim \frac{\pi}{2} - \frac{1}{x} + \frac{1}{3 x^{3}} - \frac{1}{5 x^{5}} + O \Bigl(x^{-6}\Bigr)$$

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+1 This is more intuitive to me than a chain of algebraic manipulations. In fact, for this problem, only the first order asymptotic $\arctan x = \frac{\pi}{2} - \frac{1}{x}$ suffices. – Srivatsan Dec 10 '11 at 2:51
For those who wonder how this asymptotic series was obtained, consider the identity $\arctan\,x=\frac{\pi}{2}-\arctan\frac1{x}$ and expand the arctangent on the right as a series... – J. M. Dec 10 '11 at 3:25

Let's see an elementary way:

$$\lim\limits_{x\to \infty} (2\arctan x -\pi)\ln x=\lim\limits_{x\to \infty} \arctan({\tan{(2\arctan x -\pi)})}\ln x=\lim\limits_{x\to \infty} -\arctan{\frac{2x}{x^2-1}} \ln x$$ $$=\lim\limits_{x\to \infty} \frac{-\arctan{\frac{2x}{x^2-1}}}{\frac{2x}{x^2-1}}\cdot \lim\limits_{x\to \infty}\frac{2x}{x^2-1}\ln x=-1 \cdot0=0.$$

Q.E.D.

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