# Thread: limit with composite function

1. ## limit with composite function

If $\displaystyle \displaystyle f(x) = \frac{1}{1+x}$ and $\displaystyle g(x) = \underbrace{f(f(f(x)))}_{n-times}$. Then $\displaystyle \displaystyle \lim_{n\rightarrow \infty}g(x) =$

My Solution::

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

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

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

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

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$\displaystyle \underbrace{f(f(f(x)))}_{n-times} =$

2. ## Re: limit with composite function

$\displaystyle f(x)=\frac{1}{1+x}$ and $\displaystyle g(x)=\underbrace{f(f(f(\cdots f(x) \cdots )))}_{\text{Infinite Times}}$

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

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

taking y=g(x)

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

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

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

$\displaystyle \implies g(x)=\frac{\sqrt{5}-1}{2}$

3. ## Re: limit with composite function

But Sir Here why we discard other root

4. ## Re: limit with composite function

I discarded the other root because , g(x) is clearly positive...

5. ## Re: limit with composite function

For a slightly different way of looking at it, you might notice that the n'th iterate $\displaystyle f^{(n)}(x)$ is given by $\displaystyle f^{(n)}(x) = \frac{F_nx+F_{n+1}}{F_{n+1}x + F_{n+2}},$ where $\displaystyle F_n$ is the n'th Fibonacci number. This is given by the formula $\displaystyle F_n = \bigl(\lambda_+^n-\lambda_-^n\bigr)/\sqrt5,$ where $\displaystyle \lambda_+ = \tfrac12(1+\sqrt5)$ and $\displaystyle \lambda_- = \tfrac12(1-\sqrt5).$ But $\displaystyle |\lambda_-|<1$, and so $\displaystyle \lambda_-^n\to0$ as $\displaystyle n\to\infty.$ Thus for large values of n, $\displaystyle F_n\approx \lambda_+^n/\sqrt5$ and therefore

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