The value of limn→∞ cos⁡x2cos⁡x4cos⁡x8…cos⁡x2n, is

A
1
B
$\frac{\sin x}{x}$
C
$\frac{x}{\sin x}$
D
none of these

Solution:

We know that

$\cos A \cos 2 A \cos 4 A \dots \cos 2^{n - 1} A = \frac{\sin 2^{n} A}{2^{n} \sin A}$

Taking $A = \frac{x}{2^{n}},$ we get

$\begin{array}{l} \cos (\frac{x}{2^{n}}) \cos (\frac{x}{2^{n - 1}}) \dots \cos (\frac{x}{4}) \cos (\frac{x}{2}) = \frac{\sin x}{2^{n} \sin (\frac{x}{2^{n}})} \\ ∴ lim_{n \to \infty} \cos (\frac{x}{2}) \cos (\frac{x}{4}) \dots \cos (\frac{x}{2^{n - 1}}) \cos (\frac{x}{2^{n}}) \\ = lim_{n \to \infty} \frac{\sin x}{2^{n} \sin (\frac{x}{2^{n}})} = lim_{n \to \infty} \frac{\sin x}{x} \frac{(x / 2^{n})}{\sin (x / 2^{n})} = \frac{\sin x}{x} \end{array}$

The value of $lim_{n \to \infty} \cos (\frac{x}{2}) \cos (\frac{x}{4}) \cos (\frac{x}{8}) \dots \cos (\frac{x}{2^{n}}),$ is

Solution:

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