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?
 A: $$(2 \arctan(x) - \pi)  \ln(x) = - 2 (\pi/2 - \arctan(x)) \ln(x) = -2 \arctan(1/x) \ln(x)$$
$$\lim_{x \rightarrow \infty} (2 \arctan(x) - \pi)  \ln(x) = -2 \lim_{x \rightarrow \infty} \arctan(1/x) \ln(x) = 2 \lim_{x \rightarrow 0^+} \arctan(x) \ln(x)$$
$$\lim_{x \rightarrow 0^+} \arctan(x) \ln(x) = \lim_{x \rightarrow 0^+}\frac{\arctan(x)}{\frac1{ \ln(x)}} = \lim_{x \rightarrow 0^+} \frac{\frac{1}{1+x^2}}{-\frac1{\ln^2(x)} \frac1x} = - \lim_{x \rightarrow 0^+} \frac{x \ln^2(x)}{1+x^2} = 0$$
Hence, $$\lim_{x \rightarrow \infty} (2 \arctan(x) - \pi)  \ln(x) = 0.$$
A: Maple has this asymptotic information for arctan as $x \to \infty$...
$$
\arctan x \sim \frac{\pi}{2} - \frac{1}{x} + \frac{1}{3 x^{3}} - \frac{1}{5 x^{5}} + O \Bigl(x^{-6}\Bigr)
$$
A: 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.
