1
$\begingroup$

Question: Limit of $\lim\limits_{n \to \infty} (1+ \frac{\pi}{2} - \arctan(x))^x$

The first things I notice are : $\lim\limits_{x \to \infty} \arctan(x) = \frac{\pi}{2}$ and the limit looks something of the form $(1 + \frac{1}{x})^x$. Unfortunetly, I can not seem to apply these ideas to solve the limits. I am not sure if it is right:

$$\ln(y) = \lim\limits_{x \to \infty} x \ln(1 + \frac{\pi}{2} - \arctan(x)) = \lim\limits_{x \to \infty}\frac{\ln(1+\frac{\pi}{2}- \arctan(x))}{\frac{1}{x}}$$

Apply l'hopital's rule ... ?

Could someone confirm this approach is right, or if wrong provide a correct approach?

$\endgroup$
1
  • $\begingroup$ Yes, it is right: you can try L'Hospital. $\endgroup$
    – DonAntonio
    Aug 19, 2020 at 21:17

2 Answers 2

2
$\begingroup$

Since $\frac{\pi}{2} - \arctan(x) =\arctan \left(\frac1x\right)\to 0$ we can use that

$$\left(1+ \frac{\pi}{2} - \arctan(x)\right)^x=\left[\left(1+ \arctan\left(\frac1x\right)\right)^{\frac{1}{\arctan\left(\frac1x\right)}}\right]^{x\arctan\left(\frac1x\right)}$$

and then refer to standard limits.

Or as an alternative, following your idea

$$\lim\limits_{x \to \infty}\frac{\ln(1+\frac{\pi}{2}- \arctan(x))}{\frac{1}{x}}=\lim\limits_{x \to \infty}\frac{\ln\left(1+\arctan \left(\frac1x\right)\right)}{\arctan \left(\frac1x\right)}\,\frac{\arctan \left(\frac1x\right)}{\frac1x}$$

and then conclude again by standard limits.

$\endgroup$
2
$\begingroup$

$$A=\left(1+ \frac{\pi}{2} - \tan^{-1}(x)\right)^x=\left(1+\tan ^{-1}\left(\frac{1}{x}\right)\right)^x$$

$$\log(A)=x \log\left(1+\tan ^{-1}\left(\frac{1}{x}\right)\right)$$

By Taylor $$\tan ^{-1}\left(\frac{1}{x}\right)=\frac{1}{x}-\frac{1}{3 x^3}+O\left(\frac{1}{x^5}\right)$$ $$\log\left(1+\tan ^{-1}\left(\frac{1}{x}\right)\right)=\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{12 x^4}+O\left(\frac{1}{x^5}\right)$$ $$\log(A)=1-\frac{1}{2 x}+\frac{1}{12 x^3}+O\left(\frac{1}{x^4}\right)$$ $$A=e^{\log(A)}=e \left(1-\frac{1}{2 x}+\frac{1}{8 x^2}+\frac{1}{16 x^3} \right)+O\left(\frac{1}{x^4}\right)$$

Edit

Consider $x=\frac {11}{24}\pi$ (this is quite far away from $\infty$) for which the arctangent is $\left(1+\sqrt{2}\right) \left(\sqrt{2}+\sqrt{3}\right)$.

the exact value is $1.97993$ while this truncated expression gives $1.99516$.

In fact, the relative error is smaller than $0.01$% if $x\geq3$

$\endgroup$

You must log in to answer this question.

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