Given a sequence: $$ x_{n+1} = {4\over 3}x_n - x_n^2 $$ And initial conditions: $$ x_1 = {1\over 6} $$ I want to find $\lim x_n$ as $n\to\infty$. The approach i want to use is prove $x_n$ is bounded and monotone, hence convergent and then find its fixed points one of which is going to be the limit.
I have difficulties showing $x_n$ is bounded. If I make an assumption that it's bounded then for $L = \lim x_n$: $$ 3L^2 - L = 0 \iff L = 0\ \text{or}\ L = {1\over 3} $$
Which in case $L\ne 0$ matches the answer. Could someone please assist me on showing the bounds for $x_n$? I would also like to know whether the rest reasoning is fine. Thank you.
Let's define $$e_n={1\over 3}-x_n$$then by substituting we obtain $$e_{n+1}={2\over 3}e_n+e_n^2=e_n\left({2\over 3}+e_n\right)$$with $e_1={1\over 6}$. Also note that if $e_n>0$ then $e_{n+1}>0$ therefore since $e_1>0$ we have $e_n>0$. Also if $e_n\le {1\over 6}$ we obtain $$e_{n+1}=e_n\left({2\over 3}+e_n\right)\le {5\over 6}e_n$$ finally $$0<e_{n+1}\le {5\over 6}e_n$$which means that $e_n\to 0$ and $x_n\to {1\over 3}$ therefore $x_n$ is bounded.