The value of _ n → ∞ ( cos xcos dfrac x 2 cos(x)/(4) . . . cos x 2 ^ n ) is equal to

Correct answer: C. ( 2x)/(2x) Required limit = _ n → ∞ cos x 2 ^ 0 cos x 2 ^ 1 cos x 2 ^ 2 . . . cos x 2 ^ n = _ n → ∞ 1 2sin x 2 ^ n \ cos x . . . cos x 2 ^ n - 1 (2sin x 2 ^ n cos x 2 ^ n )\ = _ n → ∞ 1 2 ^ 2 sin x 2 ^ n \ cos x . . . (2cos x 2 ^ n - 1 sin x 2 ^ n - 1 )\ = _ n → ∞ 1 2 ^ n + 1 sin x 2 ^ n (2cos xsin x) = _ n → ∞ 2x 2 ^ n + 1 ( x 2 ^ n ) = ( 2x)/(2x) _ n → ∞ ( ( x 2 ^ n ) ( x 2 ^ n ) ) = ( 2x)/(2x)

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Limits

NTA Abhyas 2020 · Mathematics · Limits

The value of _ n → ∞ ( cos xcos dfrac x 2 cos(x)/(4) . . . cos x 2...

The value of

\lim_{n \to \infty} \left(

cos $xcos\d dfrac{x}{2}cos\dfrac{x}{4} . . . cos\dfrac{x}{{2}^{n}}\right) is equal to

Options

A $\dfrac{x}{sin x}$
B $\dfrac{sin x}{x}$
C $\dfrac{\sin 2x}{2x}$
D $\dfrac{2x}{\sin 2x}$

Correct answer

C. $\dfrac{\sin 2x}{2x}$

Step-by-step solution

Required limit

= \lim_{n \to \infty}cos\dfrac{x}{{2}^{0}}cos\dfrac{x}{{2}^{1}}cos\dfrac{x}{{2}^{2}} . . . cos\dfrac{x}{{2}^{n}}

= \lim_{n \to \infty}\dfrac{1}{2sin\dfrac{x}{{2}^{n}}}\left\{cos x . . . cos\dfrac{x}{{2}^{n - 1}}\left(2sin\dfrac{x}{{2}^{n}}cos\dfrac{x}{{2}^{n}}\right)\right\}

= \lim_{n \to \infty}\dfrac{1}{{2}^{2}sin\dfrac{x}{{2}^{n}}}\left\{cos x . . . \left(2cos\dfrac{x}{{2}^{n - 1}}sin\dfrac{x}{{2}^{n - 1}}\right)\right\}

= \lim_{n \to \infty}\dfrac{1}{{2}^{n + 1}sin\dfrac{x}{{2}^{n}}}\left(2cos xsin x\right)

= \lim_{n \to \infty}\dfrac{\sin 2x}{{2}^{n + 1}\sin \left(\dfrac{x}{{2}^{n}}\right)}

= \dfrac{\sin 2x}{2x}\lim_{n \to \infty}\left(\dfrac{\left(\dfrac{x}{{2}^{n}}\right)}{\sin \left(\dfrac{x}{{2}^{n}}\right)}\right) = \dfrac{\sin 2x}{2x}

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