Limit of $x(2\pi - 4\arctan(x))$ I am trying to decide the value of the limit stated below. But I am running out of ideas, since I can't figure out how to rewrite it in a beneficial way.
$$\lim_{x \to \infty} x(2\pi - 4\arctan(x))$$

My attempt
$\lim\limits_{x \to \infty} x(2\pi - 4\arctan(x))$
rewrite $2\pi$ as $4\arctan(0)$
Substitute $x = \frac{1}{t}$
$\lim\limits_{t \to 0} \frac{1}{t}(4\arctan(t) - 4\arctan(\frac{1}{t}))$
$\lim\limits_{t \to 0} \left(\frac{4\arctan(t)}{t} - \frac{4\arctan(\frac{1}{t})}{t}\right)$
The limit of the first term is $4$, because of the standard limit $\lim\limits_{x \to 0} \left(\frac{\arctan(t)}{t}\right) = 1$. But after that I have no idea.
I would like some hints how to solve this problem, and problems similair to this. I currently have problems with limits approaching $0$.

L'Hopitals rule is more than sufficient here so this is overkill but you could also have arrived at the result by knowing the first term in the Laurent series for $\arctan(x)$ which says that
$\arctan(x) \approx \frac{\pi}{2} - \frac{1}{x} + \frac{1}{3x^3} + ...$ for large $x$. This is sort of like a Maclaurin series only with inverse powers. 
$\arctan(x)$ is deligtful because it is rather simple to derive the full formula. You start with
$$\pi/2 - \arctan(x) = \arctan(\infty) - \arctan(x) = \int_x^\infty \frac{ds}{1 + s^2} ds $$
and then make a rearrangement and expansion of the integrand with the geometric series in such a way that we get inverse powers as those are integrable to infinity
$$\frac{1}{s^2 +1} = \frac{1}{s^2}\frac{1}{1 + (-s^{-2})} = \frac{1}{s^2}\sum_{k = 0}^\infty (-1)^ks^{-2k} = \sum_{k = 0}^\infty (-1)^k \frac{1}{s^{2(k + 1)}}$$
You integrate
$$\int_x^\infty \frac{ds}{1 + s^2} ds = \sum_{k = 0}^\infty \int_{x}^\infty (-1)^k\frac{1}{s^{2(k + 1)}}ds = \sum_{k = 0}^\infty \frac{(-1)^{k}}{2k +1}\frac{1}{s^{2k + 1}}$$
and got
$$\arctan(x) = \frac{\pi}{2} + \sum_{k = 0}^\infty \frac{(-1)^{k + 1}}{2k +1}\frac{1}{s^{2k + 1}}  = \frac{\pi}{2} - \frac{1}{x} + \frac{1}{3x^3} - \frac{1}{5x^5} + ...$$
This neat formula is just a novelty though but taking the first two terms we get
$$\boxed{\lim_{x \to \infty}x (2\pi - 4\arctan(x)) =  \lim_{x \to \infty}x\left (2\pi - 4\left ( \frac{\pi}{2} - \frac{1}{x} + O\left (\frac{1}{x^3}\right )\right ) \right )}$$
$$\boxed{\lim_{x \to \infty}x (2\pi - 4\arctan(x)) = \lim_{x \to \infty} 4 + O\left ( \frac{1}{x^2}\right ) = -4 }$$
