1
$\begingroup$

I have to calculate next limit: $$\lim\limits_{x \rightarrow \infty}\left(\frac{2}{\pi}\arctan(x)\right)^{\frac{x^2}{1+2x}}$$

So far I got to this point $$e^{\lim\limits{x\rightarrow\infty}\frac{x^2}{1+2x}\frac{2\arctan(x)-\pi}{\pi}}$$ When I start to calculate this limit on $e$ I then came to this $$\frac{1}{\pi} \lim\limits_{x\rightarrow\infty}\frac{x^2(2\arctan(x)-\pi)}{1+2x}$$ And also I must not use L'Hôpital's rule for this one.

Any help?

$\endgroup$
6
  • $\begingroup$ Is it $+$ or $-\infty$, or both? $\endgroup$
    – Bernard
    Nov 29 '19 at 17:58
  • $\begingroup$ It is $+$, in this paper it is just $\infty$ $\endgroup$ Nov 29 '19 at 18:01
  • $\begingroup$ That manipulation isn't correct. You're missing a log somewhere around the arctan $\endgroup$ Nov 29 '19 at 18:01
  • $\begingroup$ Nothing is missing, because when I wrote that limit in limit calculator, it calculated with L'Hôpital's rule, so limit exist $\endgroup$ Nov 29 '19 at 18:08
  • $\begingroup$ Just because the limit exists doesn't mean you're computing the correct one. Whenever you have $f(x)^{g(x)}$ the correct manipulation is $e^{g(x)\cdot\log(f(x))}$, which you did not do. $\endgroup$ Nov 29 '19 at 18:18
1
$\begingroup$

To avoid direct L'Hopital (but not really, see my explanation below) we note that for positive $x$

$$\arctan(x) = \frac{\pi}{2} - \arctan\left(\frac{1}{x}\right)$$

And we can use this to get asymptotic approximations to the limit

$$= \lim_{x\to\infty} \left(1-\frac{2}{\pi}\arctan\left(\frac{1}{x}\right)\right)^{\frac{x}{2}-\frac{1}{4}+\frac{1}{8x+4}} \sim \lim_{x\to\infty} \left( 1 - \frac{2}{\pi x}\right)^{\frac{x}{2}} = e^{-\frac{1}{\pi}}$$

from the defintion of the limit for $e^x = \lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n$. This still uses the spirit of L'Hopital, which is asymptotic behaviors of functions.

$\endgroup$
5
  • $\begingroup$ What about the other part of limit $\frac{x^2}{1+2x}$ $\endgroup$ Nov 29 '19 at 18:25
  • $\begingroup$ @NikiSedlarevic what about it? The manipulation is there. $\endgroup$ Nov 29 '19 at 18:37
  • $\begingroup$ @user what would you recommend needs more explaining? $$\frac{x^2}{1+2x} = \frac{x}{2} - \frac{1}{4} + \frac{1}{4+8x}$$ seems like an algebraic fact that doesn't need anymore explanation. $\endgroup$ Nov 29 '19 at 19:12
  • $\begingroup$ @NinadMunshi I'm referring to the fact to avoid asymthotic using $$\left[(1+f(x))^{\frac1{f(x)}}\right]^{f(x)g(x)}$$. $\endgroup$
    – user
    Nov 29 '19 at 19:14
  • $\begingroup$ Mu point is that of course your solution is fine but the use of asymptotics (even if very effective and powerfull when handeled by an expert user) can be a little bit obscure and often can lead to erroneous evaluation for limits by less expert users. Therefore with some step more we can make things more clear in my opinion. $\endgroup$
    – user
    Nov 29 '19 at 19:19
1
$\begingroup$

We are to compute \begin{align*} \lim_{x\rightarrow\infty}\dfrac{x^{2}}{1+2x}\log\left(\dfrac{2}{\pi}\tan^{-1}x\right). \end{align*} With the change of variable $u=1-\dfrac{2}{\pi}\tan^{-1}x$, then \begin{align*} \lim_{x\rightarrow\infty}\dfrac{x^{2}}{1+2x}\log\left(\dfrac{2}{\pi}\tan^{-1}x\right)&=\lim_{u\rightarrow 0^{+}}\dfrac{\cot^{2}\dfrac{\pi}{2}u}{1+2\cot\dfrac{\pi}{2}u}\log(1-u)\\ &=\lim_{u\rightarrow 0^{+}}\dfrac{2}{\pi}\cdot\dfrac{\dfrac{\pi}{2}u}{\sin\dfrac{\pi}{2}u}\cdot\dfrac{\cos^{2}\dfrac{\pi}{2}u}{\sin\dfrac{\pi}{2}u+2\cos\dfrac{\pi}{2}u}\cdot\dfrac{1}{u}\cdot\log(1-u). \end{align*} Note that \begin{align*} \lim_{u\rightarrow 0^{+}}\dfrac{1}{u}\cdot\log(1-u)=-\lim_{u\rightarrow 0^{+}}\dfrac{1}{u}\int_{0}^{u}\dfrac{1}{1-t}dt=-\lim_{u\rightarrow 0^{+}}\dfrac{1}{1-\eta_{u}}=-1, \end{align*} where $\eta_{u}$ is in between $u$ and $0$, chosen by Mean Value Theorem.

$\endgroup$
1
  • 1
    $\begingroup$ Nice answer. My only comment is that the last is a well known standard limit derived from $(1+1/x)^x \to e$, we don't need to use MVT. $\endgroup$
    – user
    Nov 29 '19 at 19:28
0
$\begingroup$

Compute the limit of the log in the first place: $$\frac{x^2}{1+2x}\ln\tfrac\pi2+\frac{x^2}{1+2x}\ln(\arctan x).$$ The first term tends to $\dots$ as $x\to\infty$. For the second term, set $u=\frac1x$; since $x>0$, we obtain $$\frac{x^2}{1+2x}\ln(\arctan x)=\frac1{u(u+2)}\Bigl(\frac\pi2-\arctan u\Bigr)=\frac{\pi}{2u(u+2)}-\frac{\arctan u}{u(u+2)}.$$ Can you proceed now?

$\endgroup$
6
  • $\begingroup$ The first term tends to $\infty$. $\endgroup$
    – user284331
    Nov 29 '19 at 18:23
  • $\begingroup$ In the last sum, I suppose. And the second term? $\endgroup$
    – Bernard
    Nov 29 '19 at 18:24
  • $\begingroup$ No, no, I am saying that $\dfrac{x^{2}}{1+2x}\log(\pi/2)$. $\endgroup$
    – user284331
    Nov 29 '19 at 18:25
  • $\begingroup$ Oh! I see. I had in mind the exponential of this term. $+\infty$ or $-\infty$? (this is important since it is a part of the log). $\endgroup$
    – Bernard
    Nov 29 '19 at 18:27
  • $\begingroup$ @Bernard Sorry, I'm not getting the way you suggest to determine the solution! $\endgroup$
    – user
    Nov 29 '19 at 19:26
0
$\begingroup$

Since $f(x)=\frac{2}{\pi}\arctan(x)-1=-\frac 2 \pi\arctan \frac1x\to 0$ we have

$$\left(\frac{2}{\pi}\arctan(x)\right)^{\frac{x^2}{1+2x}}=\left(\left(1+f(x)\right)^{\frac1{f(x)}}\right)^{f(x)\frac{x^2}{1+2x}}\to e^{-\dfrac 1 \pi}$$

indeed

$$\left(1+f(x)\right)^{\frac1{f(x)}} \to e$$

and by standard limits

$${f(x)\frac{x^2}{1+2x}}=-\frac 2 \pi\frac{\arctan \frac1x}{\frac1x}\frac{x^2}{x+2x^2}\to-\frac2\pi\cdot 1 \cdot \frac12=-\frac1\pi$$

$\endgroup$
0
$\begingroup$

I thought it would be instructive to present a way forward that relie on only pre-calculus analysis including some elementary inequalities. To that end we now proceed.

PRIMER:

In THIS ANSWER, I showed using elementary, pre-calculus tools that the arctangent function satisfies the inequalities

$$\frac{x}{\sqrt{1+x^2}}\le \arctan(x)\le x\tag1$$

for $x\ge 0$. We now use $(1)$ in the development that follows.


Enforcing the substitution $x\mapsto 1/x$ reveals

$$\begin{align} \lim_{x\to \infty}\left(\frac2\pi \arctan(x)\right)^{\frac{x^2}{1+2x}}&\overbrace{=}^{x\mapsto 1/x}\lim_{x\to 0}\left(\frac2\pi \arctan(1/x)\right)^{\frac1{x(x+2)}}\\\\ &=\lim_{x\to 0}\left(1-\frac2\pi \arctan(x)\right)^{\frac1{x(x+2)}}\tag2 \end{align}$$


Using $(1)$ along with the inequality $\frac1{\sqrt{1+x^2}}\ge 1-\frac12 x^2$ in $(2)$, we find that

$$\left(1-\frac2\pi \left(x-\frac12x^3\right)\right)^{\frac1{x(x+2)}}\le \left(1-\frac2\pi \arctan(x)\right)^{\frac1{x(x+2)}}\le \left(1-\frac2\pi x\right)^{\frac1{x(x+2)}}\tag3$$


Next, recalling that $\lim_{x\to 0}\left(1+ tx\right)^\frac1x=e^t$, it is easy to see that $\lim_{x\to 0}\left(1-\frac2\pi x\right)^{\frac1{x(x+2)}}=e^{-1/\pi}$.

We will now show that the limit of the left-hand side of $(3)$ is also $e^{-1/\pi}$, whence application of the squeeze theorem yields the coveted limit.


Proceeding, we write the left-hand side of $(3)$ as

$$\begin{align} \left(1-\frac2\pi \left(x-\frac12x^3\right)\right)^{\frac1{x(x+2)}}&=\left(1-\frac2\pi x\right)^{\frac1{x(x+2)}}\times \color{blue}{\left(1+x\,\frac{x^2}{\pi-2 x}\right)^{\frac1{x(x+2)}}}\tag4 \end{align}$$

Using Bernoulli's Inequality, we have for $0<x<\pi$

$$\begin{align} 1\le \color{blue}{\left(1+x\,\frac{x^2}{\pi-2 x}\right)^{\frac1{x(x+2)}}}&\le \frac1{\left(1-x\,\frac{x^2}{\pi-2 x}\right)^{\frac1{x(x+2)}}}\\\\ &\le \frac1{1-\frac{x^2}{(x+2)(\pi -2x)}}\tag5 \end{align}$$

Applying the squeeze theorem to $(5)$, we find that

$$\lim_{x\to 0}\left(1+x\,\frac{x^2}{\pi-2 x}\right)^{\frac1{x(x+2)}}=1\tag6$$


Using $(6)$ in $(4)$ reveals

$$\lim_{x\to 0}\left(1-\frac2\pi \left(x-\frac12x^3\right)\right)^{\frac1{x(x+2)}}=e^{-1/\pi}\tag7$$


Finally, using $(7)$ in $(3)$ and then equating to $(2)$ yields the coveted limit

$$\lim_{x\to \infty}\left(\frac2\pi \arctan(x)\right)^{\frac{x^2}{1+2x}}=e^{-1/\pi}$$

And we are done!

Tools Used: Elementary pre-calculus analysis (e.g. Bernoulli's Inequality and other elementary inequalities) only along with the limit definition of the exponential function.

$\endgroup$
2
  • $\begingroup$ @NikiSedlarević I've posted a solution which uses only pre-calculus tools. Let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$
    – Mark Viola
    Apr 15 '20 at 22:31
  • $\begingroup$ And please feel free to accept an answer and up vote answers as you see fit. $\endgroup$
    – Mark Viola
    Apr 15 '20 at 22:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.