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)?
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Duplicate of http://math.stackexchange.com/questions/709197/how-do-i-find-the-limit-of-this-problem/709203#709203 – Michael Hoppe Mar 13 '14 at 20:56
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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 $$\lim_{x\to\frac\pi 2}\tan x=+\infty\iff \lim_{x\to+\infty}\arctan x=\frac\pi2$$
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$$\lim_{x \to +\infty} \arctan x = \frac{π}{2}$$ and $$\lim_{x \to -\infty} \arctan x = -\frac{π}{2},$$ then in total $$\lim_{x \to \infty} \arctan x = \frac{π}{2} \mathrm{sgn}(x).$$
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