# limit with composite function

• Jan 13th 2012, 11:31 PM
jacks
limit with composite function
If $\displaystyle f(x) = \frac{1}{1+x}$ and $g(x) = \underbrace{f(f(f(x)))}_{n-times}$. Then $\displaystyle \lim_{n\rightarrow \infty}g(x) =$

My Solution::

$f(x) = \frac{1}{x+1}$ Then

$f(f(x)) = \frac{x+1}{x+2}$

$f(f(f(x))) = \frac{x+2}{2x+3}$

$f(f(f(f(x)))) = \frac{2x+3}{3x+5}$

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$\underbrace{f(f(f(x)))}_{n-times} =$
• Jan 13th 2012, 11:59 PM
sbhatnagar
Re: limit with composite function
$f(x)=\frac{1}{1+x}$ and $g(x)=\underbrace{f(f(f(\cdots f(x) \cdots )))}_{\text{Infinite Times}}$

$g(x)= \frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots }}}}$

$\implies g(x)=\frac{1}{1+g(x)}$

taking y=g(x)

$\implies y(1+y)=1$

$\implies y^2+y-1=0$

$\implies y=\frac{\sqrt{5}-1}{2}$

$\implies g(x)=\frac{\sqrt{5}-1}{2}$
• Jan 15th 2012, 05:34 AM
jacks
Re: limit with composite function
But Sir Here why we discard other root
• Jan 15th 2012, 06:28 AM
sbhatnagar
Re: limit with composite function
I discarded the other root because , g(x) is clearly positive...
• Jan 15th 2012, 07:58 AM
Opalg
Re: limit with composite function
For a slightly different way of looking at it, you might notice that the n'th iterate $f^{(n)}(x)$ is given by $f^{(n)}(x) = \frac{F_nx+F_{n+1}}{F_{n+1}x + F_{n+2}},$ where $F_n$ is the n'th Fibonacci number. This is given by the formula $F_n = \bigl(\lambda_+^n-\lambda_-^n\bigr)/\sqrt5,$ where $\lambda_+ = \tfrac12(1+\sqrt5)$ and $\lambda_- = \tfrac12(1-\sqrt5).$ But $|\lambda_-|<1$, and so $\lambda_-^n\to0$ as $n\to\infty.$ Thus for large values of n, $F_n\approx \lambda_+^n/\sqrt5$ and therefore

$f^{(n)}(x) \approx\frac{\lambda_+^nx+\lambda_+^{n+1}}{\lambda _+^{n+1}x+\lambda_+^{n+2}} = \frac1\lambda_+ = \frac2{1+\sqrt5} = \tfrac12(\sqrt5-1).$