Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
The answers in this thread might help your intuition math.stackexchange.com/questions/2899/… –  AnonymousCoward Dec 21 '11 at 0:40
    
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
add comment

4 Answers

up vote 9 down vote accepted

I finally solved it with help of this picture.

enter image description here

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

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 $\arctan{(AD)} = x = \frac{\pi}{2}$.

share|improve this answer
3  
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
add comment

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 $\lim\limits_{\theta\rightarrow {\infty}}\arctan\theta={\pi\over2}$.

share|improve this answer
add comment

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,

share|improve this answer
add comment

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

enter image description here

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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