JEE MATHEMATICS MCQ: LIMITS

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Analyzing +4 | -1

Let

x_0=0, x_1=1

and

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

for

n \geq 1

.
Then

\lim \limits_{{n \rightarrow \infty}}\left(\frac{x_n}{2^{n-1}}\right)=

View answer and explanation

Correct option: D

\frac{4}{\pi}

Option D is correct. By using the recurrence relation and trigonometric identities, the limit can be evaluated to 4/π.

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