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 $0$.
 A: Be careful: $\arctan0=0$.
You can use the substitution $t=\frac{\pi}{2}-\arctan x$; for $x\to\infty$ we have $t\to0^+$; note that $\arctan x=\frac{\pi}{2}-t$, so $x=\tan(\frac{\pi}{2}-t)=\cot t$; thus the limit becomes
$$
\lim_{t\to0^+}4t\cot t=\lim_{t\to0^+}\frac{4t}{\tan t}=4
$$
Another idea could be remembering the identity
$$
\arctan t+\arctan\frac{1}{t}=\frac{\pi}{2}
$$
(for $t>0$), so trying the substitution $x=1/t$; the limit becomes
$$
\lim_{t\to0^+}\frac{2\pi-4\arctan\frac{1}{t}}{t}=
\lim_{t\to0^+}\frac{4\arctan t}{t}=4
$$
A: 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
$$\boxed{\lim_{x \to \infty}x (2\pi - 4\arctan(x)) =  \lim_{x \to \infty}x\left (2\pi - 4\left ( \frac{\pi}{2} - \frac{1}{x} + O\left (\frac{1}{x^3}\right )\right ) \right )}$$
$$\boxed{\lim_{x \to \infty}x (2\pi - 4\arctan(x)) = \lim_{x \to \infty} 4 + O\left ( \frac{1}{x^2}\right ) = -4 }$$
A: Using l'Hospital:
$$\lim_{x\to\infty}\frac{2\pi-4\arctan x}{\frac 1x}\stackrel{\text{l'H}}=\lim_{x\to\infty}\frac{-\frac4{1+x^2}}{-\frac1{x^2}}=\lim_{x\to\infty}\;4\frac{x^2}{x^2+1}=4$$
A: I solved this with L'Hopital's rule. Write it as
$$\lim\frac{2\pi-4\arctan x}{\frac1x}$$
You'll see this makes it easy.
A: $$L=\lim_{x\rightarrow \infty}x(2\pi-4\arctan(x))$$
Let $t=\frac1x$.
$$L=\lim_{t\rightarrow0}\frac{2\pi-4\arctan(\frac1t)}{t}$$
Using the L'Hopital's rule,
$$L=\lim_{t\rightarrow0}\frac{-4}{1+\frac{1}{t^2}}(\frac{-1}{t^2})=\lim_{t\rightarrow0}\frac{4}{t^2+1}=4$$
