Infinite trigonometric limit

Calculus Level 4

$\large \lim_{n \to \infty} \cos \left(\frac x2 \right) \cos \left(\frac x4 \right) \cos \left(\frac x8 \right) \cdots \cos \left(\frac x{2^n} \right)$

Find the above limit at $x=\dfrac \pi 6$

\pi

e 0

3/\pi

1 solution

A Former Brilliant Member
May 31, 2016

Multiply and divide by $2^n sin(x/2^n)$ , so the numerator will become $sinx$ {Using $2sinxcosx=sin2x$ }

therefore expression is $\frac{sinx}{2^n sin(x/2^n)}$

$\lim_{n \to \infty} \frac{sinx}{2^n sin(x/2^n)}=\frac{sinx}{\lim_{n \to \infty} 2^n sin(x/2^n)}$

as $n \to \infty$ therefore, we have $x/2^n \to 0$ therefore,

$\lim_{n \to \infty} 2^n sin(x/2^n)=x$

So, answer is $\boxed{\frac{sinx}{x}}= \frac{3}{\pi}$

Infinite trigonometric limit

1 solution

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