JEE MATHEMATICS MCQ: LIMITS

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Analyzing +4 | -1

x_{n+1}=\sqrt{\frac{1+x_n}{2}}

and

\left|x_0\right|<1, n \geq 0

, then

\lim \limits_{{n \rightarrow \infty}}\left(\frac{\sqrt{1-x_0^2}}{x_1 x_2 x_3 \ldots x_{n+1}}\right)

is equal to

View answer and explanation

Correct option: D

\cos ^{-1}\left(x_0\right)

Option D,

\cos^{-1}(x_0)

, is correct because by substituting

x_0 = \cos \theta

and using trigonometric identities, the limit simplifies to

\theta

, which is

\cos^{-1}(x_0)

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