Limit of $\arctan z$

,$\displaystyle \lim_{z \rightarrow \infty} \arctan(z) = \frac{\pi}{2}$. One way to see this is to put $\displaystyle z = \frac{y}{x}$ and imagine $y$ and $x$ as the sides of a right triangle. Then as $x$ goes to zero, $\displaystyle \frac{y}{x}$ goes to infinity and $\theta$ where $\theta = \arctan(z)$, goes to $\displaystyle \frac{\pi}{2}$. I have two questions regarding this problem:

1. Is the geometric proof above, sufficiently rigorous for higher analysis? If not, why?

2. Is there an analytic way to prove this i.e without any geometric intuition?

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And also, sorry for the delay in replying. I'd not logged in for a couple of days. – Nikhil Panikkar Mar 3 '13 at 6:36
So sorry, it was my fault. I lookedat your answers and I thought it were your questions. So sorry. I'll remove my comment. – Git Gud Mar 3 '13 at 11:29
Its ok. I'll also delete my comment. :-) – Nikhil Panikkar Mar 4 '13 at 3:38

First we prove the relation $$\forall z>0,\quad \arctan(z)+\arctan(\frac{1}{z})=\frac{\pi}{2}.$$ Indeed, let denote $f(z)=\arctan(z)+\arctan(\frac{1}{z})$ for $z>0$, then we check easily that $f'(z)=0$(use the fact that the derivative of $\arctan(z)$ is$\frac{1}{1+z^2}$), then $f$ must be a constant on the interval $(0,+\infty)$ and for $z=1$, since $\arctan(1)=\frac{\pi}{4}$ we find the result.
Now, it's easy to see that $$\lim_{z\rightarrow +\infty}\arctan z=\lim_{z\rightarrow +\infty}\frac{\pi}{2}-\arctan(\frac{1}{z})=\frac{\pi}{2}-\arctan 0=\frac{\pi}{2}.$$