<s>[INST] Find the limit of the cosine sequence. The question is to find $\lim\limits_{n\to\infty} \cos \frac{x}{2} \cdot \cos \frac{x}{2^2} \cdot \cos \frac{x}{2^3}\cdot\cdot\cdot \cos \frac{x}{2^n}$.
I first thought of several facts:
1). $-1 \leq \cos(a) \leq 1$ (From this I know the sequence is also bounded by -1 and 1)
2). $\cos(0) = 1$ (This is what the items in the sequence tend to, as n becomes bigger)
3). When $x=0$, the limit is 1.  
Then I failed to think more and the above do not get me anywhere close to finding the limit...  
Can anybody teach me how to find the limit of this sequence? Thanks a lot.
 [/INST] If $x \ne 0$, then
\begin{align}
\lim_{n\to\infty}\cos \frac{x}{2}\cos\frac{x}{2^2}\cos\frac{x}{2^3}\cdots \cos \frac{x}{2^n}&=\lim_{n\to\infty}\frac{\cos \frac{x}{2}\cos\frac{x}{2^2}\cos\frac{x}{2^3}\cdots \cos \frac{x}{2^n}\sin\frac{x}{2^n}}{\sin\frac{x}{2^n}}\\
&=\lim_{n\to\infty}\frac{\cos \frac{x}{2}\cos\frac{x}{2^2}\cos\frac{x}{2^3}\cdots \cos \frac{x}{2^{n-1}}\sin\frac{x}{2^{n-1}}}{2\sin\frac{x}{2^n}}\\
&=\lim_{n\to\infty}\frac{\cos \frac{x}{2}\cos\frac{x}{2^2}\cos\frac{x}{2^3}\cdots \cos \frac{x}{2^{n-2}}\sin\frac{x}{2^{n-2}}}{2^2\sin\frac{x}{2^n}}\\
&=\cdots\\
&=\lim_{n\to\infty}\frac{\sin x}{2^n \sin\frac{x}{2^n}}\\
&=\frac{\sin x}{x}.
\end{align}
If $x=0$, then given limit goes to $1$.
</s>