# Determine <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-TeXAtom

Determine $\underset{n\to \mathrm{\infty }}{lim}\mathrm{arctan}\left(\sqrt{n+1}\right)-\mathrm{arctan}\left(\sqrt{n}\right)$
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Julien Carrillo
Note,
$\mathrm{arctan}\sqrt{n+1}-\mathrm{arctan}\sqrt{n}=\mathrm{arctan}\frac{\sqrt{n+1}-\sqrt{n}}{1+\sqrt{n+1}\cdot \sqrt{n}}$
$=\mathrm{arctan}\frac{1}{\left(1+\sqrt{{n}^{2}+n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}$
Thus,
$\underset{n\to \mathrm{\infty }}{lim}\mathrm{arctan}\left(\sqrt{n+1}\right)-\mathrm{arctan}\left(\sqrt{n}\right)$
$=\underset{n\to \mathrm{\infty }}{lim}\mathrm{arctan}\frac{1}{\left(1+\sqrt{{n}^{2}+n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}=\mathrm{arctan}\left(0\right)=0$
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Nicholas Cruz
The arctan function is the inverse function of
$\mathrm{tan}:\left(-\frac{\pi }{2},\frac{\pi }{2}\right)\to \mathbb{R}$
as this function is monotonically increasing, we have
$\underset{x\to \frac{\pi }{2}}{lim}\mathrm{tan}x=+\mathrm{\infty }\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\underset{x\to +\mathrm{\infty }}{lim}\mathrm{arctan}x=\frac{\pi }{2}$
therefore
$\underset{n\to \mathrm{\infty }}{lim}\mathrm{arctan}\left(\sqrt{n+1}\right)=\frac{\pi }{2}$
$\underset{n\to \mathrm{\infty }}{lim}\mathrm{arctan}\left(\sqrt{n}\right)=\frac{\pi }{2}$
so the difference is 0.