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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 :

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

Any Help is Welcome. ThankYou.

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@GitGud : Man you're like my official editor now. Thanks again :) – Simar Jul 30 '13 at 23:52
Is there anything that needs clearing out that you unaccepted? – Pedro Tamaroff Jul 31 '13 at 0:10
Just wanted more answers.I'll do that in a few hours. The only thing that I couldn't get was that why does sin x/2^n come outside the product when cos was inside ? – Simar Jul 31 '13 at 0:20
Though your answer seems correct, just a silly doubt it is. Please clarify if you can. Thanks :) – Simar Jul 31 '13 at 0:21
Ah, fair enough. =) Is it clearer now? – Pedro Tamaroff Jul 31 '13 at 0:23
up vote 20 down vote accepted

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:

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

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Got that.But how do you evaluate the product of cosine series? – Simar Jul 30 '13 at 23:45
@Simar I have posted the full solution. – Pedro Tamaroff Jul 30 '13 at 23:47
@PedroTamaroff Very effective solution indeed. – Akira Apr 24 at 19:23

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 $$ \lim_{n\to\infty}u_n(x)=\lim_{\xi\to0}\frac{\sin x}{x}\cdot\left(\frac{\sin\xi}{\xi}\right)^{-1}=\frac{\sin x}{x}. $$

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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 $$\frac{\sin(x)}{x}.$$

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