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 $0$.

  • $\begingroup$ $\;2\pi\neq4\arctan 0=0\;$ , but $\;2\pi=\lim\limits_{x\to\infty}4\arctan x\;$ $\endgroup$
    – DonAntonio
    Mar 13, 2016 at 20:39
  • $\begingroup$ @ras Be careful not to mistake $\arctan$ with $\operatorname{arccot}$. $$\operatorname{arccot} 0=\frac\pi2$$ but $\arctan 0=0$. $\endgroup$
    – user228113
    Mar 13, 2016 at 20:42
  • $\begingroup$ To continue with your approach, you can use the identity: $\arctan \left( \frac{1}{t}\right)=\frac{\pi}{2}- \arctan{t}$ (got that from wikipedia: en.wikipedia.org/wiki/Inverse_trigonometric_functions#arctan). $\endgroup$ Mar 13, 2016 at 20:42

5 Answers 5


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 $$ \lim_{t\to0^+}\frac{2\pi-4\arctan\frac{1}{t}}{t}= \lim_{t\to0^+}\frac{4\arctan t}{t}=4 $$

  • $\begingroup$ Do you have any specific tips how to find a "smart" substitution? Or is it something you learn after hours of solving these types of problems? $\endgroup$
    – Ras
    Mar 13, 2016 at 21:03
  • $\begingroup$ @Ras It is often useful reducing to limits at $0$, because we know many limits of the kind. Since $\lim_{x\to\infty}\arctan x=\pi/2$, setting $\arctan x=\pi/2-t$ seems an idea to try out. I added another method more in line with the idea you pursued in the question. $\endgroup$
    – egreg
    Mar 13, 2016 at 21:15

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

$$\boxed{\lim_{x \to \infty}x (2\pi - 4\arctan(x)) = \lim_{x \to \infty}x\left (2\pi - 4\left ( \frac{\pi}{2} - \frac{1}{x} + O\left (\frac{1}{x^3}\right )\right ) \right )}$$

$$\boxed{\lim_{x \to \infty}x (2\pi - 4\arctan(x)) = \lim_{x \to \infty} 4 + O\left ( \frac{1}{x^2}\right ) = -4 }$$


Using l'Hospital:

$$\lim_{x\to\infty}\frac{2\pi-4\arctan x}{\frac 1x}\stackrel{\text{l'H}}=\lim_{x\to\infty}\frac{-\frac4{1+x^2}}{-\frac1{x^2}}=\lim_{x\to\infty}\;4\frac{x^2}{x^2+1}=4$$


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.


$$L=\lim_{x\rightarrow \infty}x(2\pi-4\arctan(x))$$ Let $t=\frac1x$. $$L=\lim_{t\rightarrow0}\frac{2\pi-4\arctan(\frac1t)}{t}$$ Using the L'Hopital's rule, $$L=\lim_{t\rightarrow0}\frac{-4}{1+\frac{1}{t^2}}(\frac{-1}{t^2})=\lim_{t\rightarrow0}\frac{4}{t^2+1}=4$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .