Recursive sequence and a quadratic equation related inequality proof I am trying to show that if a sequence of number $x_{n}$ is defined by $x_1 = h$, $x_{n+1}=x_n^2 + k$, where $0<k<\frac{1}{4}$ and $h$ lies between the roots $a$ and $b$ of the equation $$x^2 -x +k = 0$$ Then show that $$a < x_{n+1}<x_n<b$$ and i am also interested in evaluating the limit of $x_n$.
Analysis towards a solution
I suspect that geometrically this sequence may have tendencies to converge or intersect  this quadratic equation's parabola although i am unsure how to exploit this hunch. What else do I know $$x^2 -x +k = 0 = (x-a)(x-b)$$ hence $a + b =1$ and $0 < k = ab < \frac{1}{4}$
Although i am unsure how to proceed from here any help would be much appreciated.
 A: Let us try to rescue one more damsel-question from the unanswered questions dragon's den. 
First, the roots of the quadratic equation $\,x^2-x+k=0\,$ are $$\frac{1\pm\sqrt{1-4k}}{2}\Longrightarrow a:=\frac{1-\sqrt{1-4k}}{2} < h< \frac{1+\sqrt{1-4k}}{2}=:b$$
Observing the geometric interpretation of the above, we have the upwards parabola $\,f(x)=x^2-x+k\,$ with two intersection points with the $\,x-$axis, both with positive abscissa, and such that $f(h)=h^2-h+k<0\,$ , since $\,f(x_0)<0 \Longleftrightarrow a<x_0<b\,$ .
Clearly $\,f(x_1)=f(h)<0\,$, and we also have 
$$x_2:=x_1^2+k<x_1\Longleftrightarrow f(x_1)=x_1^2-x_1+k<0\Longleftrightarrow a<x_1=h<b $$
Thus, we see that, in general, 
$$a<x_n<b\Longleftrightarrow f(x_n)=x_n^2-x_n+k<0\Longleftrightarrow x_{n+1}:=x_n^2+k<x_n$$
So it is enough to prove now inductively on the index of $\,\{x_n\}\,$ that $\,x_{i+1}>x_i\,\,,\,\forall i\in\mathbb N\,$ ; assuming for $\,i< n\,$ we prove it for $\,i=n$: 
$$x_{n+1}=x_n^2+k<x_n\Longleftrightarrow f(x_n)=x_n^2-x_n+k\stackrel {ind. hypothesis}<0$$
Thus, $\,\{x_n\}\,$ is a monotonically decreasing sequence bounded below by $\,a\,$ , so its limit exists, call it $\,\alpha\,$. Using arithmetic of limits and the recursion $\,x_{n+1}=x_n^2+k\,$ we get 
$$\alpha\xleftarrow [\infty\leftarrow n]{}{\color{red} {x_{n+1}=x_n^2+k}}\xrightarrow [n\to\infty]{} \alpha+k\,\,\Longrightarrow \alpha^2-\alpha+k=0\Longrightarrow \alpha=a$$ since the sequence decreases.
