Limit of Product!

Consider the product $P(n)$ as follows:

$\begin{aligned} P(n) & = \prod_{k=1}^n \cos \frac x{2^k} \\ P({\color{#D61F06}1}) & = \cos \frac x2 = \frac {\sin \frac x2}{\sin \frac x2} \times \cos \frac x2 = \frac {\sin x}{2^{\color{#D61F06}1} \sin \frac x{2^{\color{#D61F06}1}}} \\ P({\color{#D61F06}2}) & = \cos \frac x2 \cos \frac x4 = \frac {\sin x}{2\sin \frac x2} \times \cos \frac x4 = \frac {\sin x}{4\sin \frac x4 \cos \frac x4}\times \cos \frac x4 = \frac {\sin x}{2^{\color{#D61F06}2} \sin \frac x{2^{\color{#D61F06}2}}} \end{aligned}$

We can readily prove by induction that $P({\color{#D61F06}n}) = \dfrac {\sin x}{2^{\color{#D61F06}n} \sin \frac x{2^{\color{#D61F06}n}}}$ . Therefore, we have:

$\large \begin{aligned} L & = \lim_{n \to \infty} \prod_{k=1}^n \cos \frac x{2^k} \\ & = \lim_{n \to \infty} \frac {\sin x}{2^n \sin \frac x{2^n}} \\ & = \lim_{n \to \infty} \frac {\sin x}{\frac {x\sin \frac x{2^n}}{\frac x{2^n}}} & \small \color{#3D99F6} \text{Let }u = \frac x {2^n} \\ & = \lim_{u \to 0} \frac {\sin x}{x \frac {\sin u}u} \\ & = \boxed{\dfrac {\sin x}x} \end{aligned}$

By the theorem of Euler we know that Sin x = x cos x /2 . cos x/2^2.........till infinity.. Just now cross multiplying x we get our answer. And the Euler theorem can also be proved easily by simply using sine double angle formulae and some approximation as sinx is approximately equal to x for very small angles.