# Why is $\lim\limits_{x \space \to \infty}\space{\arctan(x)} = \frac{\pi}{2}$?

As part of this problem, after substitution I need to calculate the new limits.

However, I do not understand why this is so:

$$\lim_{x \to \infty}\space{\arctan(x)} = \frac{\pi}{2}$$

I tried drawing the unit circle to see what happens with $\arctan$ when $x \to \infty$ but I don't know how to draw $\arctan$. It is the inverse of $\tan$ but do you even draw $\tan$?

I would appreciate any help.

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Since the limit you ask for is also just the limit of this integral. Either way, these questions should be linked for peoples' reference. – AnonymousCoward Dec 21 '11 at 0:43

I finally solved it with help of this picture.

• $\sin x = BC$
• $\cos x = OB$
• $\tan x = AD$
• $\cot x = EF$
• $\sec x = OD$
• $\csc x = OF$

Note that, our nomenclature of $\tan x$ is not really arbitrary. $AD$ is really the tangent to the unit circle at A. Now it is clearly visible that when $\tan{(x)} = AD \to \infty$ then $\arctan{(AD)} = x = \frac{\pi}{2}$.

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Of course it is not arbitrary! :D – Mariano Suárez-Alvarez Dec 21 '11 at 2:52
The reason why $\tan(x) = AD$ is because of Thales : $\sin x / \cos x = (\tan x)/1$. So of course I agree with Mariano :P – Patrick Da Silva Jun 5 '13 at 0:16

If you wanted to do it geometrically, your proof is the easiest. If you wanted to do it analytically, you can use the fact that $\mathrm{\tan} : (-\pi/2,\pi/2) \to \mathbb R$ is an increasing continuous bijection (even an homeomorphism), thus $$\lim_{x \to \infty} \mathrm{arctan}(x) = \lim_{y \to \pi/2} \mathrm{arctan}(\mathrm{tan}(y)) = \lim_{y \to \pi/2} y = \pi/2.$$ In other words you just do the change of variables.

Hope that helps,

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Here's a slightly different way of seeing that $\lim\limits_{\theta\rightarrow {\infty}}\arctan\theta={\pi\over2}$.

Thinking of the unit circle, $\tan \theta ={y\over x}$, where $(x,y)$ are the coordinates of the point on the unit circle with reference angle $\theta$, what happens as $\theta\rightarrow\pi/2$? In particular, what happens to $\tan\theta$ as $\theta\nearrow{\pi\over2}$?

Well, the $x$ coordinate heads towards 0 and the $y$ coordinate heads towards 1.

So in the quotient $$y\over x,$$ the numerator heads to 1 and the denominator becomes arbitrarily small; so the quotient heads to infinity.

Thus, $\lim\limits_{\theta\rightarrow {\pi\over2}}\tan\theta=\infty$ and consequently $\lim\limits_{\theta\rightarrow {\infty}}\arctan\theta={\pi\over2}$.

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Since you mentioned the picture of $y =\arctan x$, have you looked it up in Wikipedia?

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