<s>[INST] Solve $\ell=\lim _{x \rightarrow \frac{\pi}{2}} \frac{(1-\sin x)\left(1-\sin ^{2} x\right) \ldots\left(1-\sin ^{n} x\right)}{\cos ^{2 n} x}$ $\ell=\lim _{x \rightarrow \frac{\pi}{2}} \frac{(1-\sin x)\left(1-\sin ^{2} x\right) \ldots\left(1-\sin ^{n} x\right)}{\cos ^{2 n} x}$
I have tried pinning it between $ \frac{(1-\sin x)^n}{\cos ^{2 n} x}$ and $ \frac{(1-\sin^nx)^n}{\cos ^{2 n} x}$ but the limits differed. Tried substituting ${x}$ for ${y}={x}-\pi/2$, no results. The derivative seems too complicated to use L'Hopital, so I'm out of ideas.
 [/INST] We take the limit $\lim\limits_{x \to \frac{\pi}{2}} \frac{1 - \sin^n{x}}{\cos^2{x}}$. We can evaluate this by $\lim\limits_{x \to \frac{\pi}{2}} \frac{1 - \sin^n{x}}{\cos^2{x}} = \lim\limits_{x \to \frac{\pi}{2}} \frac{1 - \sin^n{x}}{1 - \sin^2{x}} = \lim\limits_{u \to 1} \frac{1 - u^n}{1 - u^2}$, which we evaluate to $\frac{n}{2}$ by L'Hopital's rule.
So $\lim\limits_{x \to \frac{\pi}{2}} \frac{(1 - \sin{x}) \cdot ... \cdot (1 - \sin^n{x})}{\cos^{2n}{x}} = \lim\limits_{x \to \frac{\pi}{2}} \prod\limits_{j = 1}^n \frac{1 - \sin^j{x}}{\cos^2 {x}} = \prod\limits_{j = 1}^n \lim\limits_{x \to \frac{\pi}{2}} \frac{1 - \sin^j{x}}{\cos^2 {x}} =
\prod\limits_{j = 1}^n \frac{j}{2} = \frac{\prod\limits_{j = 1}^n j}{\prod\limits_{j = 1}^n 2} = \frac{n!}{2^n}$.
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