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.
 A: 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,
A: 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 
$\lim\limits_{\theta\rightarrow {\infty}}\arctan\theta={\pi\over2}$.
A: 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 $\arctan{(AD)} = x = \frac{\pi}{2}$.
A: Since you mentioned the picture of $y =\arctan x$, have you looked it up in Wikipedia?

