Show a sequence converges, and determine its limit 
I have a sequence $\langle x_{n}\rangle$ is defined by $x_1 = h$ and $x_{n+1} = x^2_n+k$ where $0<k<1/4$ and $h$ lies between the roots $a$ and $b$ of the equation $x^2-x+k = 0$
(1) Prove $a<x_{n+1}\le x_n<b$ where $n=1,2,3,\ldots\,$.
(2) Show $\langle x_n\rangle$ converges and determine its limit

My take when I first saw this question, is that I wanted to prove $(1)$ by triangle inequality, but I don't think I can apply that here. So I used the $a<x_n<b$ by induction.
Let $$n=1, a < x_1 < b,$$ so $n=1$ is true
Let $$n=n+1, a < x_{n+1} < b \Longrightarrow a < x^2_n + k < b,$$
but I'm stuck in this process. How do I further continue proving with induction?
Also, could I get some general hint or steps for how to start (2)?
all I could think of for (2) is that, $X_n \to L$ as $n$ goes to infinity
 A: If you prove (1) then your sequence is monotonic descending and bounded below by $\;a\;$ , thus it is convergent, say to $\;L\;$, and then by arithmetic of limits:
$$L\xleftarrow[\infty\leftarrow n]{}x_{n+1}=x_n^2+k\xrightarrow[n\to\infty]{}L^2+k\implies L^2-L+k=0,\,\text{and thus}\;\;L=a\;$$
S0 proving (1) solves all: the roots of $\;x^2-x+k=0\;$ are
$$x_{1,2}=\frac{1\pm\sqrt{1-4k}}2\implies a=\frac{1-\sqrt{1-4k}}2\;,\;\;b=\frac{1+\sqrt{1-4k}}2\implies$$
$$a<x_1=h<b$$
Inductively: $\;x_2=h^2+k\le h=x_1\iff h^2-h+k\le0$ , which is true since the parabola 
$\;f(x)=x^2-x+k\;$ is an upwards vertical one and is thus non-positive for any value between its 
roots, and since $\;h\;$ is one such value, we get $\;f(h)\le0\;$ . Also observe that
$$a\le x_2= h^2+k\iff h^2-a+k\ge0 \,.\; \text{But}\;\;h^2-a+k\ge a^2-a+k=0 $$
since $\;h\ge a\;$ . Similarly, $\;x_2<b\;$
Now induction:
$$x_{n+1}=x^2_n+k\le x_n\iff x_n^2-x_n+k\le 0$$
and the claim follows by induction both on monotony and on $\;a<x_k<b\;$
