How to calculate $\lim_{x \rightarrow \infty} (4x\arctan(x)-2\pi x)$

Question

How to calculate $$\lim_{x \rightarrow \infty} (4x\arctan(x)-2\pi x)$$

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I checked Wolframalpha and the answer is -4, shall I use some kinda standard limit? I dont get anywhere.

Answer 1

$$\begin{align}\lim_{x\to\infty}(4x\arctan(x)-2\pi x)&=\lim_{x\to\infty}\frac{\arctan(x)-\frac{\pi}{2}}{\frac{1}{4x}}\\&=\lim_{x\to\infty}\frac{\left(\arctan(x)-\frac{\pi}{2}\right)'}{\left(\frac{1}{4x}\right)'}\\&=\lim_{x\to\infty}\frac{\frac{1}{x^2+1}}{-\frac{1}{4x^2}}\\&=\lim_{x\to\infty}-4\cdot\frac{1}{1+\frac{1}{x^2}}\\&=-4.\end{align}$$

Answer 2

$$4x\arctan x-2\pi x=-4x\left(\frac\pi2-\arctan x\right)=-4x\cdot\text{arccot}(x)$$

Setting $\dfrac1x=h$

$$\lim_{x \rightarrow \infty} (4x\arctan(x)-2\pi x)=-4\lim_{h\to0^+}\frac{\text{arccot}(1/h)}h=-4\lim_{h\to0^+}\frac{\arctan h}h=?$$

See also : Simplifying an Arctan equation