Question:
If \(\lim_{n \to \infty} x_n\) exists and is finite, \(x_1 = 2\), \(x_{n+1} = \frac{a + b x_n}{b + c x_n}\) \(\forall n \in \mathbb{N}\) and \(c \gt b \gt a \gt 0\), then \(\lim_{n \to \infty} x_n =\)
If \(\lim_{n \to \infty} x_n\) exists and is finite, \(x_1 = 2\), \(x_{n+1} = \frac{a + b x_n}{b + c x_n}\) \(\forall n \in \mathbb{N}\) and \(c \gt b \gt a \gt 0\), then \(\lim_{n \to \infty} x_n =\)
Show Hint
For linear recurrence \(x_{n+1} = \frac{a + bx_n}{b + c x_n}\), set limit \(L\) and solve \(L = \frac{a + bL}{b + cL}\) to find \(\lim x_n\).
Updated On: Jul 18, 2026
- \(\sqrt{\frac{ab}{c}}\)
- \(\sqrt{\frac{a}{c}}\)
- \(\sqrt{\frac{c}{b}}\)
- \(\sqrt{\frac{a}{b}}\)
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The Correct Option is B
Solution and Explanation
Step 1: Set limit.
Let \(\lim_{n \to \infty} x_n = L\). Then \[ L = \frac{a + b L}{b + c L} \]
Step 2: Solve for L.
\[ L(b + c L) = a + b L \implies bL + c L^2 = a + b L \implies c L^2 = a \implies L = \sqrt{\frac{a}{c}} \]
Step 3: Check positivity.
Given all parameters positive, \(L \gt 0\)
Step 4: Final conclusion.
Hence, \[ \boxed{\sqrt{\frac{a}{c}}} \]
Let \(\lim_{n \to \infty} x_n = L\). Then \[ L = \frac{a + b L}{b + c L} \]
Step 2: Solve for L.
\[ L(b + c L) = a + b L \implies bL + c L^2 = a + b L \implies c L^2 = a \implies L = \sqrt{\frac{a}{c}} \]
Step 3: Check positivity.
Given all parameters positive, \(L \gt 0\)
Step 4: Final conclusion.
Hence, \[ \boxed{\sqrt{\frac{a}{c}}} \]
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