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 :  
$$
\lim \limits_{n \to \infty}
\left[\cos\left(x \over 2\right)\cos\left(x \over 4\right)
      \cos\left(x \over 8\right)\ \cdots\ \cos\left(x \over 2^{n}\right)\right]
$$

Hint $$\begin{align}{\sin x}&=2^1\sin\frac x 2 \cos\frac x2\\{}\\\sin x& =2^2\sin \frac x4\cos\frac x 4\cos \frac x 2\\{}\\\sin x& =2^3\sin \frac x8\cos \frac x8\cos\frac x 4\cos \frac x 2\\{}\\\cdots\;&=\hspace{2cm }\cdots\end{align} $$
One further hint

$$\sin x = {2^n}\sin \frac{x}{{{2^n}}}\prod\limits_{k = 1}^n {\cos \frac{x}{{{2^k}}}} $$

You'll need $\dfrac{\sin x}x\to 1$ as $x\to 0$.
Final spoiler:

$$\lim\limits_{n \to \infty } \prod\limits_{k = 1}^n \cos {\frac{x}{2^k}}  = \lim \limits_{n \to \infty } \frac{\sin x}{x}\left( \frac{\sin {2^{ - n}x}}{2^{ - n}x} \right)^{ - 1} = \frac{\sin x}{x}$$

