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$\lim\limits_{x\to \infty}\cos(\large\frac{x}{2})$$\cos(\large\frac{x}{4})$$\cos(\large\frac{x}{8})$$........\cos(\large\frac{x}{2^n})$ is

$\begin{array}{1 1}(a)\;1&(b)\;\large\frac{\sin x}{x}\\(c)\;\large\frac{x}{\sin x}&(d)\;\text{None of these}\end{array}$

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Let $\large\frac{x}{2^n}$$=\alpha$
$\Rightarrow x=2^n x$
As $n\to \infty\qquad \alpha \to 0$
$\lim\limits_{\alpha\to 0}\sin\alpha=\alpha$-------(1)
Now from trigonometry
$\cos\alpha\cos 2\alpha.....\cos n\alpha=\large\frac{\sin(2^n\alpha)}{2^n\sin \alpha}$(By (1))
$\Rightarrow \large\frac{\sin x}{x}$
Hence (b) is the correct answer.
answered Jan 6, 2014 by sreemathi.v

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