real analysis - Show that ${{x_n}}$ is convergent and monotone

Question: For $c>0$, consider the quadratic equation

$$
x^2-x-c=0,\qquad x>0.
$$
Define the sequence $\{x_n\}$ recersively by fixing $x_1>0$ and then, if $n$ is an index for which $x_n$ has been defined, defining
$$
x_{n+1}=\sqrt{c+x_n}.
$$
Prove that the sequence $\{x_n\}$ converges monotonically to the solution of the above equation.

My uncompleted solution: General speaking, the sequence ${\{x_{n+1}}\}$ is a subsequence of ${\{x_n}\}$. Hence, if $\lim_{n \to \infty} x_n = x_s$, then $\lim_{n \to \infty} x_{n+1} = x_s$ as well. So, from sum and productproperties of convergent sequences,

(finally) it follows that $x_s=\sqrt{c+x_s}$ which is equivalent to the mentioned quadratic equation for $x>0$. To show that ${\{x_n}\}$ is monotone, it is enough to show that it is bounded, since ${\{x_n}\}$ is convergent. But I don't know how to show that ${\{x_n}\}$ is bounded.

Thank you in advance for a clear guidance/solution.

EDIT : (by considering first two comments to this question, so) The question is, show that ${\{x_n}\}$ is $1-$ convergent, and, $2-$ monotone.

Answer

Let $f(x)=\sqrt{c+x}$, $x>0$. There is a unique fixed point $x^*$ such that $f(x^*)=x^*$. $x^*$ is in fact the positive solution of $x^2-x-c=0$. The sequence $x_n$ is defined by $x_{n+1}=f(x_n)$.

If $0

If $x^*x>x^*$. From this it is easy to show that if $x_1>x^*$ then $x_n$ is decreasing and bounded below by $x^*$. This implies that $x_n$ converges and the limit is $x^*$.

If $x_1=x^*$, then $x_n=x^*$ for all $n$.