Limit of the nested radical $x_{n+1} = \sqrt{c+x_n}$ (Fitzpatrick Advanced Calculus 2e, Sec. 2.4 #12)
For $c \gt 0$, consider the quadratic equation
$x^2 - x - c = 0, x > 0$.
Define the sequence $\{x_n\}$ recursively by fixing $|x_1| \lt c$ 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.
Note: The answers below might assume $x_1 \gt 0$, but they still work, as we have $x_3 \gt 0$.
 A: Let $k$ be the positive root to your polynomial. Note that $y=x^2-x-c$ is an upward opening parabola with its vertex below the $x$-axis and an initial downward slope. This implies that positive $x$-values less than $k$ produce negative output, while $x$-values greater than $k$ produce positive output.
Note also that all $x_n$ are positive, so it will be acceptable to preserve equalities and inequalities involving $x_n^2$ after taking a square root.
If $x_0=k$, then $x_1^2=c+k=k^2$, so $x_1=k$. The sequence continues like this, and is constant.
If $x_n<k$, then $x_{n+1}^2=c+x_n<c+k=k^2$. So $x_{n+1}<k$. (Similarly if $x_n>k$, then $x_{n+1}>k$.) This establishes that the sequence is either bounded above or below, depending on where $x_0$ is in relation to $k$. 
If $x_n<k$, then $x$ is a positive number to the left of the root of your polynomial. $x$-values in this region produce negative output, so $x_n^2-x_n-c<0$. That implies that $x_{n+1}^2=c+x_n>x_n^2$, and so $x_{n+1}>x_n$.  (Similarly if $x_n>k$, then $x_{n+1}<x_n$.) 
Thus if $x_0<k$ you will have an increasing sequence bounded above. And if $x_0>k$ you will have a decreasing sequence bounded below. 
So the limit exists under all possible cases. It's value has to be a solution to $L=\sqrt{c+L}$. There is only one such solution: $L=k$.
A: Let $r=\frac {1+\sqrt {1+4c}}2$ be the positive root of the quadratic, so that $r^2=r+c$ and $r\gt 1$
Note that for $n\gt 1$ we have $x_n\gt 0$
Now suppose $r\gt x_n$ so with $x_{n+1}^2=c+x_n$ we have $$r^2-x_{n+1}^2=(r+c)-(c+x_n)=r-x_n\gt 0$$and  $$r-x_{n+1}=\frac {r-x_n}{r+x_{n+1}}\lt r-x_n$$
Whence $x_n$ is monotonically increasing, and getting closer to $r$ - the difference reduces at least as fast as $r^{-n}$, so the limit is easy to prove.
On the other hand if $r\lt x_n$ we have $$x_{n+1}^2-r^2=(c+x_n)-(r+c)=x_n-r\gt 0$$and  $$x_{n+1}-r=\frac {x_n-r}{x_{n+1}+r}\lt x_n-r$$and the sequence is decreasing and bounded below by $r$, and it is once again easy to prove that this is the limit.
A: Assuming that you know that a monotone, bounded sequence converges, you want to do two things. First, show that $\langle x_n:n\in\mathbb{Z}^+\rangle$ is monotone and bounded, and then show that its limit is the positive root of $x^2-x-c=0$.
If $c=x_1=1$, $x_2=\sqrt2>x_1$, while if $c=1$ and $x_1=2$, $x_2=\sqrt3<x_1$, so if the sequence is monotonic, the direction in which it’s monotonic must depend on $c$ and $x_1$. A good first step would be to try to figure out how this dependence works.
The positive root of the quadratic is $\frac12(1+\sqrt{1+4c})$, which I’ll denote by $r$. If $x_n\to r$, as claimed, and does so monotonically, it must be the case that the sequence increases monotonically if $x_1<r$ and decreases monotonically if $x_1>r$. In the examples in the last paragraph, $r=\frac12(1+\sqrt5)\approx 1.618$, so they behave as predicted.
This suggests that your first step should be to show that if $x_n<r$, then $x_n<x_{n+1}<r$, while if $x_n>r$, $x_n>x_{n+1}>r$; that would be enough to show that $\langle x_n:n\in\mathbb{Z}^+\rangle$ is both monotone and bounded and hence that it has a limit.
Suppose that $0\le x_n<r$; you can easily check that $x_n^2-x_n-c<0$, i.e., that $x_n^2<x_n+c$. On the other hand, $x_{n+1}^2=c+x_n$, so $x_{n+1}^2>x_n^2$, and therefore $x_{n+1}>x_n$. Is it possible that $x_{n+1}\ge r$? That would require that $x_{n+1}^2-x_{n+1}-c\ge 0$ (why?) and hence that $$x_{n+1}^2\ge x_{n+1}+c>x_n+c=x_{n+1}^2\;,$$ which is clearly impossible. Thus, if $0\le x_n<r$, we must have $x_n<x_{n+1}<r$, as desired. I leave the case $x_n>r$ to you.
Once this is done, you still have to show that the limit of the sequence really is $r$. Let $f(x)=\sqrt{c+x}$; clearly $f$ is continuous, so if the sequence converges to $L$, we have $$L=\lim_{n\to\infty}x_n=\lim_{n\to\infty}x_{n+1}=\lim_{n\to\infty}f(x_n)=f(L)\;,$$ and from there it’s trivial to check that $L=r$.
Added: Note that although the problem gave us $x_1>0$, this isn’t actually necessary: all that’s needed is that $x_1\ge -c$, so that $x_2$ is defined, since $x_2=\sqrt{c+x_1}\ge 0$ automatically.
A: Hint on methodology:
Denoting $f(x)=\sqrt{c+x}$ the function that defines the recursive sequence, you can show the sequence is increasing (resp.decreasing) by showing that all its terms live in an interval $I$ such that
$$f(x)>x \quad(\text{resp.}\;f(x)<x)\quad\forall x\in I. $$
To show that  all terms of the sequence live in $I$, you just have to show that $f(I)\subseteq I$.
Last point: all terms are obviously positive. On the other hand, as $f$ is continuous, if a limit $\ell$ exists, it satisfies the equation
$$f(x)=\sqrt{c+x}=x\iff x^2-x-c=0,\quad x>0$$
This quadratic equation has a negative and a positive root: $\;\dfrac{1\pm \sqrt{1+4c}}2$. Therefore, if the sequence converges, its limit  is the positive root $\lambda$, and you can try to show that
$$f\bigl([0,\lambda]\bigr)\subseteq[0,\lambda]\quad\text{and}\quad f(x)>x\enspace\text{on }\; [0,\lambda].$$
A: I am going to do this Ramanujan-style: pick some real positive $a$.
$$
a=\sqrt{a^2}=\sqrt{a^2-a+a}=\sqrt{a^2-a+\sqrt{a^2-a+a}}=\sqrt{a^2-a+\sqrt{a^2-a+\sqrt{a^2-a+a}}}=\dots=\sqrt{a^2-a+\sqrt{a^2-a+\sqrt{a^2-a+\sqrt{a^2-a+\dots}}}}
$$
What you do is keep replacing the last $a$ in the expression by $\sqrt{a^2-a+a}$.
Now let $a^2-a=c$ and we have the given expression. Solve that for $a$ and you have your answer.
