While working on part probability question, I had actually to advice $\lim_x \to \infty \arctan(x)$. I knew the price intuitively together $\pi/2$, yet i cannot figure out exactly how to prove that by elementary method (without resorting to $\epsilon-\delta$ arguments). How does one prove the (preferably, without resorting to L"Hopital"s rule)?

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The $\arctan$ function is the inverse duty of $$\tan:\left(-\frac\pi2,\frac\pi2\right)\rightarrow\Bbb R$$and since this role is monotonically increasing then$$\lim_x\to\frac\pi 2\tan x=+\infty\iff \lim_x\to+\infty\arctan x=\frac\pi2$$

$$\lim_x \to +\infty \arctan x = \fracπ2$$ and also $$\lim_x \to -\infty \arctan x = -\fracπ2,$$then in total$$\lim_x \to \infty \arctan x = \fracπ2 \surfacetoairnewyork.comrmsgn(x).$$



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