Finding the limit $\lim \limits_{n \to \infty}\ (\cos \frac x 2 \cdot\cos \frac x 4\cdot \cos \frac x 8\cdots \cos \frac x {2^n}) $ This limit seemed quite unusual to me as there aren't any intermediate forms or series expansions which are generally used in limits. Stuck on this for a while now .Here's how it goes :  
$$
\lim \limits_{n \to \infty}
\left[\cos\left(x \over 2\right)\cos\left(x \over 4\right)
      \cos\left(x \over 8\right)\ \cdots\ \cos\left(x \over 2^{n}\right)\right]
$$

Put $z=e^{i x/2^n}.$ Then the product becomes
$$\prod_{k=1}^n \cos\left(\frac{x}{2^k}\right)
= \frac{1}{2^n} \prod_{k=0}^{n-1} \left(z^{2^k}+z^{-2^k}\right)
= \frac{1}{2^n} z^{-\sum_{k=0}^{n-1} 2^k} 
\prod_{k=0}^{n-1} \left(z^{2^{k+1}}+1\right)\\
= \frac{1}{2^n} z^{1-2^n} \frac{1}{z+1} \prod_{k=0}^n \left(z^{2^k}+1\right).$$
Now the product in this last formula is easily seen to be the generating function of the positive integers less than $2^{n+1}-1$ (consider the binary representation of an integer $q$ from this interval), so that we may continue with
$$\frac{1}{2^n} z^{1-2^n} \frac{1}{z+1} \left(1+z+z^2+\cdots+z^{2^{n+1}-1}\right)
= \frac{1}{2^n} z^{1-2^n} \frac{1}{z+1} \frac{1-z^{2^{n+1}}}{1-z}\\
= \frac{1}{2^n} \frac{1}{z+1} \frac{z^{1-2^n}-z^{1+2^n}}{1-z}
= \frac{1}{2^n} \frac{z}{z+1} \frac{z^{-2^n}-z^{2^n}}{1-z}
= \frac{1}{2^n} z \frac{z^{-2^n}-z^{2^n}}{1-z^2} \\
= \frac{1}{2^n} \frac{z^{-2^n}-z^{2^n}}{1/z-z}
= \frac{1}{2^n}  \frac{\sin(x)}{\sin(x/2^n)}
= \frac{x/2^n}{\sin(x/2^n)} \frac{\sin(x)}{x}.$$
Finally recall that $\sin(x) \sim x$ in a neighborhood of zero, so that the limit of the first term is one, resulting in a final answer of
$$\frac{\sin(x)}{x}.$$
