Limits, Continuity & Differentiability
Limits
Grade 12
Question:
<p>If \(f(n+1) = \dfrac{1}{2}\left\{f(n) + \dfrac{9}{f(n)}\right\}\) where \(n \in N\) and \(f(n) > 0\ \forall n \in N\) and \(\underset{n\to\infty}{\text{Lt}}\ f(n)\) exist then the value of \(\underset{n\to\infty}{\text{Lt}}\ f(n) =\)</p>
Step-by-Step Solution
Key Concept: If the limit exists as n→∞, then both f(n) and f(n+1) converge to the same value L. Substitute this limiting value into the recurrence relation to find L algebraically.
<p><strong>Step 1:</strong> Since lim(n→∞) f(n) exists, let lim(n→∞) f(n) = L where L > 0.</p><p><strong>Step 2:</strong> As n→∞, both f(n) and f(n+1) approach L. Taking the limit of both sides of the recurrence relation:<br>L = (1/2){L + 9/L}</p><p><strong>Step 3:</strong> Solve for L:<br>2L = L + 9/L<br>L = 9/L<br>L² = 9<br>L = 3 (since f(n) > 0)</p><p><strong>Step 4:</strong> Verify: f(n+1) = (1/2){3 + 9/3} = (1/2){3 + 3} = 3 ✓</p><p>∴ Answer: <strong>3</strong></p>
Correct Answer: 3