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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?

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  • $\begingroup$ Is it $+$ or $-\infty$, or both? $\endgroup$
    – Bernard
    Nov 29, 2019 at 17:58
  • $\begingroup$ It is $+$, in this paper it is just $\infty$ $\endgroup$ Nov 29, 2019 at 18:01
  • $\begingroup$ That manipulation isn't correct. You're missing a log somewhere around the arctan $\endgroup$ Nov 29, 2019 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, 2019 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, 2019 at 18:18

5 Answers 5

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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.

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  • $\begingroup$ What about the other part of limit $\frac{x^2}{1+2x}$ $\endgroup$ Nov 29, 2019 at 18:25
  • $\begingroup$ @NikiSedlarevic what about it? The manipulation is there. $\endgroup$ Nov 29, 2019 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, 2019 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, 2019 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, 2019 at 19:19
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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.

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    $\begingroup$ 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, 2019 at 19:28
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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?

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  • $\begingroup$ The first term tends to $\infty$. $\endgroup$
    – user284331
    Nov 29, 2019 at 18:23
  • $\begingroup$ In the last sum, I suppose. And the second term? $\endgroup$
    – Bernard
    Nov 29, 2019 at 18:24
  • $\begingroup$ No, no, I am saying that $\dfrac{x^{2}}{1+2x}\log(\pi/2)$. $\endgroup$
    – user284331
    Nov 29, 2019 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, 2019 at 18:27
  • $\begingroup$ @Bernard Sorry, I'm not getting the way you suggest to determine the solution! $\endgroup$
    – user
    Nov 29, 2019 at 19:26
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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

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

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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.

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  • $\begingroup$ @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, 2020 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, 2020 at 22:31

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