2022-03-25

Compute $\underset{x\to \frac{\pi }{2}}{lim}\frac{\left(1-\mathrm{sin}x\right)\left(1-{\mathrm{sin}}^{2}x\right)\dots \left(1-{\mathrm{sin}}^{n}x\right)}{{\mathrm{cos}}^{2n}x}$

Chelsea Chen

$\underset{x\to \frac{\pi }{2}}{lim}\frac{\left(1-\mathrm{sin}x\right)\left(1-{\mathrm{sin}}^{2}x\right)\dots \left(1-{\mathrm{sin}}^{n}x\right)}{{\mathrm{cos}}^{2n}x}=\underset{x\to \frac{\pi }{2}}{lim}\frac{1-\mathrm{sin}x}{{\mathrm{cos}}^{2}x}\cdot \frac{1-{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}\cdot \frac{1-{\mathrm{sin}}^{n}x}{{\mathrm{cos}}^{2}x}$
$=\frac{1}{2}\cdot \frac{2}{2}\dots \frac{n}{2}=\frac{n!}{{2}^{n}}$
because, near beahaves like $\frac{k}{2}{\left(x-\frac{\pi }{2}\right)}^{2}$ whereas ${\mathrm{cos}}^{2}x$ behaves like ${\left(x-\frac{\pi }{2}\right)}^{2}$

cineworld93uowb

Note that
$\underset{x\to \frac{\pi }{2}}{lim}\frac{1-{\mathrm{sin}}^{k}\left(x\right)}{1-\mathrm{sin}\left(x\right)}=\underset{x\to \frac{\pi }{2}}{lim}\sum _{j=0}^{k-1}{\mathrm{sin}}^{j}\left(x\right)$
=k
and
$\underset{x\to \frac{\pi }{2}}{lim}\frac{{\mathrm{cos}}^{2n}\left(x\right)}{{\left(1-\mathrm{sin}\left(x\right)\right)}^{n}}=\underset{x\to \frac{\pi }{2}}{lim}{\left(1+\mathrm{sin}\left(x\right)\right)}^{n}$
$={2}^{n}$
Thus, after dividing the numerator and denominator by ${\left(1-\mathrm{sin}\left(x\right)\right)}^{n}$ we get
$\underset{x\to \frac{\pi }{2}}{lim}\frac{\prod _{k=1}^{n}\left(1-{\mathrm{sin}}^{k}\left(x\right)\right)}{{\mathrm{cos}}^{2n}\left(x\right)}=\frac{n!}{{2}^{n}}$

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