Suppose f(n+1)=12f(n)+9f(n),n∈N. If f(n)>0,∀n∈N, then Ltn→∞f(n)=

Ltn→∞f(n)=Ltn→∞f(n+1)=xf(n+1)=12f(n)+9f(n) ⇒x=12x+9x ⇒2x2=x2+9 ⇒x2=9⇒x=±3 ⇒limn→∞f(n)=3

Suppose f(n+1)=12f(n)+9f(n),n∈N. If f(n)>0,∀n∈N, then Ltn→∞f(n)=

Ltn→∞f(n)=Ltn→∞f(n+1)=xf(n+1)=12f(n)+9f(n) ⇒x=12x+9x ⇒2x2=x2+9 ⇒x2=9⇒x=±3 ⇒limn→∞f(n)=3

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Suppose $f (n + 1) = \frac{1}{2} \{f (n) + \frac{9}{f (n)}\}, n \in N$ . If $f (n) > 0, \forall n \in N$ , then $\underset{n \to \infty}{L t} f (n) =$

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Detailed Solution

$\underset{n \to \infty}{L t} f (n) = \underset{n \to \infty}{L t} f (n + 1) = x$
$f (n + 1) = \frac{1}{2} \{f (n) + \frac{9}{f (n)}\} \Rightarrow x = \frac{1}{2} (x + \frac{9}{x}) \Rightarrow 2 x^{2} = x^{2} + 9 \Rightarrow x^{2} = 9 \Rightarrow x = \pm 3 \Rightarrow \lim_{n \to \infty} f (n) = 3$