Find all $a$ such that $\{x_n\}$ has finite limit Let $\{x_n\}$ be a real sequence defined by:
$$
x_1=a \\
x_{n+1}=3x_n^3-7x_n^2+5x_n
$$
For all $n=1,2,3...$ and $a$ is a real number.
Find all $a$ such that $\{x_n\}$ has finite limit when $n\to +\infty$ and find the finite limit in that cases.
My thought:
The answer is $a\in [0,\frac{4}{3}]$
we find root of $f(x)=x$
 A: Plot in the same figure the two curves
$$y=x\ ,\qquad y=f(x):=3x^3-7x^2+5x\ .$$
You then will see that $f(x)=x$ for $x=0, \ 1,\ {4\over3}$. Furthermore $f'(0)=5$, $f'(1)=0$, and $f'\bigl({4\over3}\bigr)={7\over3}$. This shows that $1$ is an attracting fixpoint of $f$ whereas the other two fixpoints are repelling. But we can learn more from the picture: If $x_0=a<0$ or $x_0=a>{4\over3}$ then obviously $\lim_{n\to\infty} x_n=-\infty$, resp. $=+\infty$.
When $x_0=a\in \ \bigl]0,{4\over3}\bigr[\ $ then things are more complicated. The function $f$ has a local maximum at $x={5\over9}$, but $f\bigl({5\over9}\bigr)={275\over243}<{4\over3}$, and a local minimum at the fixpoint $1$. This implies that in any case $$f\Bigl(\bigl]0,{4\over3}\bigr[\Bigr)\subset\bigl]0,{4\over3}\bigr[\ ,$$
so that an initial point $x_0\in \bigl]0,{4\over3}\bigr[\ $ can never escape to one of $\pm\infty$.
Note that $f\bigl({1\over3}\bigr)=1$. When $x_n<{1\over3}$ then $x_n<f(x_n)=x_{n+1}<1$, so the sequence is increasing until the first time $x_n\geq{1\over3}$. When ${1\over3}\leq x_n\leq1$ then $1\leq f(x_n)\leq {275\over243}$. When $1< x_n<{4\over3}$ then $1< x_{n+1}= f(x_n)< x_n<{4\over3}$; therefore the $x_n$ will now be monotonically decreasing to $1$. Altogether it follows that for $x_0=a\in\bigl]0,{4\over3}\bigr[\ $ we get $\lim_{n\to\infty} x_n=1$.
