# Get the limit from <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-

Get the limit from
$\underset{x\to \mathrm{\infty }}{lim}\left(x{\int }_{\frac{\pi }{2}}^{\mathrm{arctan}\left(x\right)}\mathrm{sin}\left({t}^{2}\right)dt\phantom{\rule{mediummathspace}{0ex}}\right)$
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hofyonlines5
You have
$\begin{array}{rl}\underset{x\to \mathrm{\infty }}{lim}x{\int }_{\pi /2}^{\mathrm{arctan}x}\mathrm{sin}\left({t}^{2}\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t& =\underset{x\to \mathrm{\infty }}{lim}\frac{{\int }_{\pi /2}^{\mathrm{arctan}x}\mathrm{sin}\left({t}^{2}\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t}{\frac{1}{x}}\\ & =\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{1}{1+{x}^{2}}\mathrm{sin}\left({\mathrm{arctan}}^{2}x\right)}{-\frac{1}{{x}^{2}}}\\ & =-\underset{x\to \mathrm{\infty }}{lim}\frac{{x}^{2}}{1+{x}^{2}}\mathrm{sin}\left({\mathrm{arctan}}^{2}x\right)\\ & =-\mathrm{sin}\left(\frac{{\pi }^{2}}{4}\right).\end{array}$