Compute $\lim_{x\to\frac{\pi}{2}} \frac{(1-\sin x)(1-\sin^2x)\dots(1-\sin^nx)}{\cos^{2n}x}$ 
$$\lim_{x\to\frac{\pi}{2}} \frac{(1-\sin x)(1-\sin^2x)\dots(1-\sin^nx)}{\cos^{2n}x}=?$$

Here's what I have done so far:
$y=\frac{\pi}{2}-x$ The limit becomes:$$\lim_{y\to 0}\frac{(1-\cos y)(1-\cos^2y)\dots(1-\cos^ny)}{\sin^{2n}y}=\lim_{y\to 0}\frac{1-\cos y}{\sin^2y}\frac{1-\cos^2y}{\sin^2y}\dots\frac{1-\cos^ny}{\sin^2y}$$
Also $1-\cos y=2\sin^2 \frac{x}{2}$
How should i write $1-\cos^2y ,1-\cos^3y,\dots1-\cos^ny$  in order to get the final answer. 
 A: \begin{align}\lim_{x\to\frac\pi2}\frac{(1-\sin x)(1-\sin^2x)\dots\lim(1-\sin^nx)}{\cos^{2n}x}&=\lim_{x\to\frac\pi2}\frac{1-\sin x}{\cos^2x}\cdot\frac{1-\sin^2x}{\cos^2x}\cdot\frac{1-\sin^nx}{\cos^2x}\\&=\frac12\cdot\frac22\cdots\frac n2=\frac{n!}{2^n}\end{align}because, near $\frac\pi2$, $1-\sin^kx$ beahaves like $\frac k2\left(x-\frac\pi2\right)^2$, whereas $\cos^2x$ behaves like $\left(x-\frac\pi2\right)^2$.
A: Note that
$$
\begin{align}
\lim_{x\to\pi/2}\frac{1-\sin^k(x)}{1-\sin(x)}
&=\lim_{x\to\pi/2}\sum_{j=0}^{k-1}\sin^j(x)\\
&=k
\end{align}
$$
and
$$
\begin{align}
\lim_{x\to\pi/2}\frac{\cos^{2n}(x)}{(1-\sin(x))^n}
&=\lim_{x\to\pi/2}\left(1+\sin(x)\right)^n\\[6pt]
&=2^n
\end{align}
$$
Thus, after dividing the numerator and denominator by $(1-\sin(x))^n$ we get
$$
\lim_{x\to\pi/2}\frac{\prod\limits_{k=1}^n\left(1-\sin^k(x)\right)}{\cos^{2n}(x)}=\frac{n!}{2^n}
$$
A: Note that
$$1-cos^ky=1-(1-\frac{y^2}{2}+o(y^2))^k=\frac{k}{2}y^2+o(y^2)$$
thus
$$\frac{(1-\cos y)(1-\cos^2y)\dots(1-cos^ny)}{\sin^{2n}y}=\frac{\frac{n!}{2^n}y^{2n}+o(y^{2n})}{y^{2n}+o(y^{2n})}=\frac{\frac{n!}{2^n}+o(1)}{1+o(1)}\to \frac{n!}{2^n}$$
A: Look into this
$\lim_{x\rightarrow 0} \frac{(1-\cos x)}{\sin^2 x} \cdot \frac{1-\cos^2 x}{\sin^2 x} \cdots \frac{1-\cos^n x}{\sin^2 x}$ the lim distribute over the fractions and each can be calculated using Lohpital rule
