Task 2: Solutions
Using the Algebra of Limits, show that if $(x_n)$ converges, then it must converge to $0$, or $1-1/a$.
Suppose $(x_n)$ converges with limit $x$. Then $\lim (x_{n+1})$ also equals $x$. And, by the Algebra of Limits, $\lim (x_{n+1})$ equals $\lim (a \cdot x_n \cdot (1-x_n)) = a \cdot (\lim (x_n)) \cdot (1 - \lim (x_n)) = a \cdot x \cdot (1-x)$. Solving $x = a \cdot x \cdot (1-x)$ gives $x=0$ or $x=1-1/a$.
For $0< a <1$, show that $(x_n)$ is decreasing, and converges to $0$ (for any seed).
Fix $a \in (0,1)$ and $x_1 \in [0,1]$. As shown in Task 1, the sequence $(x_n)$ is bounded by $0$ and $1$ (because $f : [0,1] \to [0,1]$). Further, $|x_{n+1}| = a \cdot x_n \cdot (1-x_n) \le a \cdot x_n$, and hence, by induction $|x_{n+1}| \le a^n \cdot x_1$. Since $0 < a < 1$, $\lim (a^n) =0$, and so $\lim (x_n) = \lim (x_{n+1}) =0$.
n+1}) =0$.