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

Setting
$$
u_n(x)=\cos\frac{x}{2}\cdot\cos\frac{x}{4}\cdot\cdots\cos\frac{x}{2^n},
$$
we have
$$
v_n(x):=u_n(x)\cdot\sin\frac{x}{2^n}=u_{n-1}(x)\cdot\cos\frac{x}{2^n}\cdot\sin\frac{x}{2^n}=\frac12u_{n-1}(x)\cdot\sin\frac{x}{2^{n-1}}=\frac12v_{n-1}(x).
$$
It follows that
$$
v_n(x)=\frac{1}{2^{n-1}}v_1(x)=\frac{1}{2^{n-1}}\cdot\cos\frac{x}{2}\cdot\sin\frac{x}{2}=\frac{1}{2^n}\sin x.
$$
Hence, provided $\sin\frac{x}{2^n} \ne 0$, we have
$$
u_n(x)=\frac{\sin x}{2^n\sin\frac{x}{2^n}}.
$$
Thus
$$
\lim_{n\to\infty}u_n(x)=\lim_{\xi\to0}\frac{\sin x}{x}\cdot\left(\frac{\sin\xi}{\xi}\right)^{-1}=\frac{\sin x}{x}.
$$
