[tex]\lim\limits_{x\to \infty}x(\pi-2\arctan x)= \lim\limits_{x\to \infty}\dfrac{\pi-2\arctan x}{\dfrac{1}{x}} =\\ \\ \overset{L'Hopital\,\,(\frac{0}{0})}{=} \lim\limits_{x\to \infty}\dfrac{0-\dfrac{2}{1+x^2}}{-\dfrac{1}{x^2}}=\lim\limits_{x\to \infty}\dfrac{2x^2}{x^2+1} = \boxed{2}[/tex]
