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I saw this question on the site. It asked to evaluate:

$$\lim_{x\to \infty}\left(\frac{2\arctan(x)}{\pi}\right)^x.$$

Although the answer is $e^{\tfrac{-2}{\pi}}$, I don't completely understand why the limit is not equal to zero. I think it should be zero because as $x\to\frac{\pi}2^-$, $\tan x\to+\infty$. Hence we have the limit of $(\frac2{\pi}\times{(\frac{\pi}2}^-))^{\infty}$. and it should be zero (for example the value of $0.99999^{1000000}$ is very close to zero.

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  • $\begingroup$ $0.99999^{1000000}$ may be close to $0$, but $0.9999999^{1000000}$ is close to $1$. A limit that looks like $(1^-)^\infty$ can wind up anywhere between $0$ and $1$. $\endgroup$ Mar 23, 2021 at 17:42

4 Answers 4

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Note that you can not just plug the limit it, since the form you get is indeterminate: $1^{\infty}$. To show you another (very famous) example why this does not work:

Consider $$\lim_{n\to\infty}(1+\frac{1}{n})^n.$$ You may recall that this limit is the defintion of $e\approx2.7$. But if we were to plug in the limit immediatly we'd get $$(1+0)^\infty.$$ And $e\neq0$ (nor $1$ or $\infty$).


Same thing for $$\lim_{x\to \infty}\left(\frac{2\arctan(x)}{\pi}\right)^x.$$ We also have $1^\infty$, but as you said correctly it's $e^{-2/\pi}$.


Addendum: This post might be helpful.

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  • $\begingroup$ Thank you for the answer. the link also was very useful, it changed my perspective of limits. $\endgroup$
    – Soheil0098
    Mar 23, 2021 at 18:24
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The statement

As $x \to \frac{\pi}{2}^-$, $\arctan x \to +\infty$

is incorrect. You are confusing $\tan x$ and $\arctan x$.

Second, your suggestion that the indeterminate form $1^\infty \to 0$ is also incorrect. For instance,

$$\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e,$$ in which $1 + 1/x \to 1$ as $x \to \infty$. However,

$$\lim_{x \to \infty} \left(1 + \frac{1}{x^2}\right)^x = 1,$$ yet $1 + 1/x^2 \to 1$ also as $x \to \infty$. And

$$\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^{x^2} = \infty.$$ All of these are indeterminate forms of type $1^\infty$ yet they have different limits, none of which are zero.

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The result highly depends on the "speed" with which $\frac{2}{\pi} \arctan x$ tends to $1$ as $x$ tends to infinity . You cannot just come up with an example like $0.99999^{1000000}$ and expect that the value has any connection with the value of the limit... If that was the case, there would be no indeterminations: every limit of the type $1^{\infty}$ would be zero.

The appropriate approximation here would be

$$ 0.9999993633802277^{1000000}\approx 0.529078 \approx e^{-2/\pi} $$

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  • $\begingroup$ @Soheil0098 Yes, $\lim_{x\to +\infty} \frac{2}{\pi} \arctan x = 1$. $\endgroup$ Mar 23, 2021 at 18:26
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Write

$$\left(\frac{2\arctan x}\pi\right)^x=e^{x\log\frac{2\arctan x}x}$$

and now L'hospital

$$\lim_{x\to\infty}\frac{\log\frac{2\arctan x}\pi}{\frac1x}\stackrel{\text{L'H}\frac00}=\lim_{x\to\infty}\frac{\frac1\pi\frac2{1+x^2}\frac\pi{2\arctan x}}{-\frac1{x^2}}=\lim_{x\to\infty}-\frac{ x^2}{1+x^2}\cdot\frac1{\arctan x}=-1\cdot\frac1{\frac\pi2}=-\frac2\pi$$

and by continutiyt of the exponential function, the limit is $\;e^{-2/\pi}\;$

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