f1(x)=x/2+10 fn(x)=f1(fn-1(x)). n>=2 Then evaluate. Lim fn(x). n tends to infinity.

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f1(x)=x/2+10 fn(x)=f1(fn-1(x)). n>=2 Then evaluate. Lim fn(x). n tends to infinity.

Mathematics
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is this what you said? \[f_1(x)=\frac{x}{2}+10 \\ f_n(x)=f_1(f_{n-1}(x)) , n \ge 2 \] I would find the first few terms and see if I can find a pattern for an explicit form for f_n
\(f_\color{red}{2}(x) = f_1*f_{2-1}= f_1^\color{red}{2}\) \(f_\color{red}{3}(x) = f_1*f_{3-1}= f_1^\color{red}{3}\) \(f_\color{red}{4}(x) = f_1*f_{4-1}= f_1^\color{red}{4}\) -------------------------------- \(f_\color{red}{n}(x) = f_1*f_{2-1}= f_1^\color{red}{n }=(\dfrac{x+20}{2})^n\)
the middle term of the last line is wrong, it should be \(f_1f_{n-1} \)

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@Loser66 I disagree: \[f_1(x)=\frac{x+20}{2}~~\implies~~f_2(x)=\frac{\dfrac{x+20}{2}+20}{2}=\frac{x+60}{4}\neq\left(\frac{x+20}{2}\right)^2\]
I believe you're mistaking composition for multiplication: \[f(f(x))\neq f(x)\times f(x)\]
oh yeah!! I misread the problem. :) Thanks for pointing it out.
I'm wondering if there's a way to do this with generating functions... Here's what I have so far. Denote \(a_1=f_1(x)\) and \(a_n=f_n(x)\), so we have the recurrence relation \[\begin{cases}a_1=\dfrac{x}{2}+10\\\\ a_n=\dfrac{a_{n-1}}{2}+10&\text{for }n\ge2\end{cases}\] Then denote the generating function by \(F(y)=\displaystyle\sum_{n=1}^\infty a_ny^n\). We have \[\begin{align*} a_n&=\frac{a_{n-1}}{2}+10\\\\ \sum_{n=2}^\infty a_ny^n&=\frac{1}{2}\sum_{n=2}^\infty a_{n-1}y^n+10\sum_{n=2}^\infty y^n\\\\ F(y)-a_1y&=\frac{y}{2}\sum_{n=2}^\infty a_{n-1}y^{n-1}+\frac{10y^2}{1-y}\\\\ \left(1-\frac{y}{2}\right)F(y)&=\frac{10y^2}{1-y}+a_1y\\\\ F(y)&=20-2a_1+\frac{20}{1-y}+\frac{20-4a_1}{2-y}\\\\ &=20-2a_1+20\sum_{n=0}^\infty y^n+(10-2a_1)\sum_{n=0}^\infty\left(\frac{y}{2}\right)^n \end{align*}\] but I'm just not seeing where to go from here...
the first 5 terms are: \(f_1 = \frac{x + 20}{2}\) \(f_2 = \frac{x + 60}{4}\) \(f_1 = \frac{x + 140}{8}\) \(f_1 = \frac{x + 300}{16}\) \(f_1 = \frac{x + 620}{32}\) this demands: \(f_n(x) = \frac{x + (2^n - 1)20}{2^n}\) which you can stuff back into the recursion to get \(f_{n+1}(x)\). and so for the limit we use \(f_n = \frac{\frac{x}{2^n} + (1 - \frac{1}{2^n})20}{1}\). 20.

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