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The $\arctan$ role is the inverse duty of $$\tan:\left(-\frac\pi2,\frac\pi2\right)\rightarrow\Bbb R$$and since this duty is monotonically enhancing 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 $$\lim_x \to -\infty \arctan x = -\fracπ2,$$then in total$$\lim_x \to \infty \arctan x = \fracπ2 \bromheads.tvrmsgn(x).$$
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