2. Answer: 1) $x_{n}=\frac{3\left(2^{n}-1\right)}{2^{n}}$ b 2) $\lim _{n \rightarrow \infty} x_{n}=3$

Option 0

For each natural number $n_{\text {write a formula for solving } x_{n}}$ of the equation $F_{n}(x)=\underbrace{f(f(f \ldots(f(x)) \ldots)}_{n}=0$, where the function $F_{n}(x)$ is the $n-$ fold composition of the function $f(x)=a x+b,|a|>1$. For example, $F_{2}(x)=f(f(x))$ and so on. Find $\lim _{n \rightarrow \infty} x_{n}$.
Answer: 1) $\left.x_{n}=-\frac{b\left(a^{n}-1\right)}{a^{n}(a-1)}, 2\right) \lim _{n \rightarrow \infty} x_{n}=-\frac{b}{a-1}$

Solution.

$$
\begin{aligned}
& F_{1}(x)=a x+b, F_{2}(x)=a(a x+b)+b=a^{2} x+b(a+1) \\
& F_{3}(x)=a\left(a^{2} x+b(a+1)\right)+b=a^{3} x+b\left(a^{2}+a+1\right), \ldots \\
& F_{n}(x)=a^{n} x+b\left(a^{n-1}+a^{n-2}+\ldots+1\right)=a^{n} x+b \cdot \frac{a^{n}-1}{a-1}
\end{aligned}
$$

The equation $F_{n}(x)=0$ has the solution $x_{n}=-\frac{b\left(a^{n}-1\right)}{a^{n}(a-1)}$. For $|a|>1$, the limit $\lim _{n \rightarrow \infty} x_{n}=-\frac{b}{a-1}$ exists and is finite.