While functioning on some probability question, I had to advice \$lim_x o infty arctan(x)\$. I knew the prize intuitively together \$pi/2\$, yet i cannot number out how to prove that by elementary way (without resorting come \$epsilon-delta\$ arguments). How does one prove it (preferably, there is no resorting to L"Hopital"s rule)?

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The \$arctan\$ role is the inverse function of \$\$ an:left(-fracpi2,fracpi2 ight) ightarrowBbb R\$\$and due to the fact that this function is monotonically boosting then\$\$lim_x ofracpi 2 an x=+inftyiff lim_x o+inftyarctan x=fracpi2\$\$

\$\$lim_x o +infty arctan x = fracπ2\$\$ and also \$\$lim_x o -infty arctan x = -fracπ2,\$\$then in total\$\$lim_x o infty arctan x = fracπ2 eer-selection.comrmsgn(x).\$\$

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