How to prove that limit of arctan(x) as x tends to infinity, is $\pi/2$? While working on some probability question, I had to evaluate $\lim_{x \to \infty} \arctan(x)$. I knew the answer intuitively as $\pi/2$, yet I cannot figure out how to prove it by elementary means (without resorting to $\epsilon-\delta$ arguments). How does one prove it (preferably, without resorting to L'Hopital's rule)?
The $\arctan$ function is the inverse function of $$\tan:\left(-\frac{\pi}2,\frac{\pi}2\right)\rightarrow\Bbb R$$
and since this function is monotonically increasing then
$$\lim_{x\to\frac\pi 2}\tan x=+\infty\iff \lim_{x\to+\infty}\arctan x=\frac\pi2$$
$$\lim_{x \to +\infty} \arctan x = \frac{π}{2}$$ 
and 
$$\lim_{x \to -\infty} \arctan x = -\frac{π}{2},$$
then in total
$$\lim_{x \to \infty} \arctan x = \frac{π}{2} \mathrm{sgn}(x).$$