<s>[INST] 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?
$$
\lim_{n\to\infty}\cos\frac{x}{2}\cos\frac{x}{4}\dotsm\cos\frac{x}{2^{n}}
$$ 
 [/INST] 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 $$ \lim_{h\to 0}\frac{\sin(h) }{h}  = 1$$
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