Let $ x_1=1 \quad \text { and } \quad x_{n+1}=\frac{4+3 x_n}{3+2 x_n} \text { for } n \geq 1 $ If $\lim _{n \rightarrow \infty} x_n$ exists finitely, then the limit is equal to

JEE MATHEMATICS MCQ: LIMITS

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Application +4 | -1

Let

x_1=1 \quad \text { and } \quad x_{n+1}=\frac{4+3 x_n}{3+2 x_n} \text { for } n \geq 1

\lim _{n \rightarrow \infty} x_n

exists finitely, then the limit is equal to

View answer and explanation

Correct option: A

\sqrt{2}

Option A (

\sqrt{2}

) is correct because the limit, if it exists, must satisfy the recursive relation, leading to the equation

l = \frac{4+3l}{3+2l}

, which solves to

l = \sqrt{2}

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