Desirae Washington

2022-07-13

How do I evaluate the limit
$\underset{x\to 0}{lim}⌊\frac{\mathrm{tan}x\mathrm{sin}x}{{x}^{2}}⌋$

Elijah Benjamin

Expert

Using Taylor–Young expansions: we know that
$\mathrm{tan}\left(x\right)=x+\frac{{x}^{3}}{3}+o\left({x}^{4}\right),\phantom{\rule{2em}{0ex}}\mathrm{sin}\left(x\right)=x-\frac{{x}^{3}}{6}+o\left({x}^{4}\right),$
hence
$\mathrm{tan}\left(x\right)\mathrm{sin}\left(x\right)={x}^{2}+\frac{{x}^{4}}{6}+o\left({x}^{5}\right),$
and hence
$\frac{\mathrm{tan}\left(x\right)\mathrm{sin}\left(x\right)}{{x}^{2}}=1+\frac{{x}^{2}}{6}+o\left({x}^{3}\right).$
From here, we conclude that there exists a punctured neighborhood V of 0 such that

Since
$\underset{x\to 0}{lim}\frac{\mathrm{tan}\left(x\right)\mathrm{sin}\left(x\right)}{{x}^{2}}=1$
we can assume that V has been chosen such that

Hence

from which we conclude that
$\underset{x\to 0}{lim}⌊\frac{\mathrm{tan}\left(x\right)\mathrm{sin}\left(x\right)}{{x}^{2}}⌋=1.$

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