# Why the $\lim_{x\to \infty}\left(\frac{2\arctan(x)}{\pi}\right)^x$ is not equal to $0$?

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.

• $0.99999^{1000000}$ 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 $1$. Mar 23, 2021 at 17:42

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 $$e^{-2/\pi}$$.

• Thank you for the answer. the link also was very useful, it changed my perspective of limits. Mar 23, 2021 at 18:24

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.

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

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

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

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 $$\;e^{-2/\pi}\;$$