<s>[INST] Prove that $x_{n+1}=x_n+0.8x_n(1-x_n)-0.072$ converges Assume that $x_1$ is a real number and for $n \gt 1$ :
$x_{n+1}=x_n+0.8x_n(1-x_n)-0.072$  
a) Assume that $x_1=0.6 $
Prove that $\{x_n\}$ converges and find its limit.  
b) Find a set of real numbers like $A$ such that if $x_1 \in A$ then $\{x_n\}$ is convergent.   
Note 1 : I know a method to answer the part "a" but its not a precise way. ( Put $L$ instead of every $x_i$ )
Note 2 : I have no idea about part "b".
 [/INST] Let 
\begin{align}
f(x) &= x+0.8x(1-x)-0.072 \\
 &= x+\frac{4}{5}x(1-x) - \frac{9}{125} \\
 &= \frac{9}{10}+\frac{9}{25}\left(x-\frac{9}{10}\right)-\frac{4}{5} \left(x-\frac{9}{10}\right)^2 
\end{align}
I obtained that last expression by expanding $f$ in a series about $9/10$. You can expand out the last two expressions to see that they are equal. 
Subtracting $9/10$ off from both sides, factoring out an $(x-9/10)$ from the right and taking the absolute values, we get
\begin{align}
  \left|f(x)-\frac{9}{10}\right| &= \left|x-\frac{9}{10}\right|
\left|\frac{9}{25}-\frac{4}{5}\left(x-\frac{9}{10}\right)\right|.
\end{align}
Thus, the distance between $f(x)$ and $9/10$ will be less than the distance between $x$ and $9/10$ precisely when
$$\left|\frac{9}{25}-\frac{4}{5}\left(x-\frac{9}{10}\right)\right| < 1.$$
This last inequality is equivalent to 
$$\frac{1}{10}<x<\frac{13}{5}.$$
Thus, the sequence will converge to $9/10$ whenever $x$ is in that interval.
</s>