<p>If \(a_1 = 1\) and \(a_{n+1} = \dfrac{4 + 3a_n}{3 + 2a_n}\), \(n \geq 1\) and if \(\lim_{n \to \infty} a_n = a\), then the value of \(a\) is</p>
Step-by-Step Solution
Key Concept: At the limit, the recurrence relation becomes a fixed point equation where a = f(a). Since a_n → a, we also have a_{n+1} → a, allowing us to substitute both sides with the same limit value.
<p><strong>Step 1:</strong> Since $\lim_{n \to \infty} a_n = a$ exists, we also have $\lim_{n \to \infty} a_{n+1} = a$ (both approach the same limit).</p><p><strong>Step 2:</strong> Taking the limit of both sides of the recurrence relation:</p><p>$$a = \frac{4 + 3a}{3 + 2a}$$</p><p><strong>Step 3:</strong> Cross-multiply to get the fixed point equation:</p><p>$$a(3 + 2a) = 4 + 3a$$</p><p>$$3a + 2a^2 = 4 + 3a$$</p><p>$$2a^2 = 4$$</p><p>$$a^2 = 2$$</p><p>$$a = \pm\sqrt{2}$$</p><p><strong>Step 4:</strong> Determine the valid solution. Starting with $a_1 = 1 > 0$, we can verify that all terms remain positive: $a_2 = \frac{4+3}{3+2} = \frac{7}{5} > 0$. By induction, $a_n > 0$ for all $n$, so $a = \sqrt{2}$.</p><p>∴ Answer: $a = \sqrt{2}$ (Option A)</p>
Correct Answer: A