Find, $ \lim_{n\to\infty}\cos\frac{x}{2}\cos\frac{x}{4}\dotsm\cos\frac{x}{2^{n}} $ What's the answer is this limit and how is it solved?
$$
\lim_{n\to\infty}\cos\frac{x}{2}\cos\frac{x}{4}\dotsm\cos\frac{x}{2^{n}}
$$ 
 A: Hint: Multiply - divide by $$2\sin \frac x{2^n}$$
Edit:
Consider the product $$L=\prod_\limits{i=1}^n\cos\frac x{2^i}$$
Then $$L=\frac 12\frac{\sin\left(\dfrac x{2^n}\right)}{\sin\left(\dfrac x{2^n}\right)}\cos\left(\frac x{2^n}\right)\ldots\cos\left(\frac x2\right)$$
$$\Rightarrow L=\frac 1{\sin \left(\dfrac x{2^n}\right)}\frac 1{2^2}*2\sin\left(\frac x{2^{n-1}}\right)\cos \left(\frac x{2^{n-1}}\right)\ldots\cos\left(\frac x2\right)$$ and so on.    
I hope this is enough to proceed further!
A: Check by induction that
or $$ \sin(x)=2\sin(\frac {x}{2^{}})\cos(\frac {x}{2^{}})=2^{2}\sin(\frac {x}{2^{2}})\cos(\frac {x}{2^{2}})\cos(\frac {x}{2^{}})\\\\ =2^{3}\sin(\frac {x}{2^{3}})\cos(\frac {x}{2^{3}}) \cos(\frac {x}{2^{2}})\cos(\frac {x}{2^{}}) =....=2^{n}\sin(\frac {x}{2^{n}})  \prod_{j=1}^n \cos(\frac {x}{2^{j}} x)$$
that is 
$$\sin(x)=2^{n}\sin(\frac {x}{2^{n}})  \prod_{j=1}^n \cos(\frac {x}{2^{j}} x) $$
So $$\prod_{j=1}^n \cos(\frac {x}{2^{j}} x)  = \frac{\sin x}{2^{n}\sin(\frac {x}{2^{n}}) } $$
Let set $h =\frac {x}{2^{n}}\to 0~~as~~n\to \infty $ then 
$$\lim_{n\to \infty }\prod_{j=1}^n \cos(\frac {x}{2^{j}} x)  = \lim_{n\to \infty }\frac{\sin x}{x } \frac{x}{2^{n}\sin(\frac {x}{2^{n}}) } \\=\frac{\sin x}{x } \lim_{h\to 0}\frac{h}{\sin(h) } =\frac{\sin x}{x }$$
since $$ \lim_{h\to 0}\frac{\sin(h) }{h}  = 1$$
