What is $\lim\limits_{n \to \infty} x_n$ where $x_1=2$ and $x_{n+1}=123+\sqrt{4+5 \sin(x_n) + 6\sqrt{x_n}}$? $$x_{n+1}=123+\sqrt{4+5 \sin(x_n) + 6\sqrt{x_n}}, \quad x_1=2$$
At first glance, this sequence seems like it will diverge, since it seems like every term is growing by at least $123$.
However, I know that this sequence converges (it is a problem from a math competition) to a number between $131$ and $132$, but I have no idea how to calculate it. 

This is a more subtle problem than most.
First, consider the equation $t = 123 + \sqrt {9 + 6 \sqrt t}$ for $t \ge 0$. This equation has a unique solution $A \ge 0$; to see that, note that the LHS is less than the RHS for $t=0$ but tends to infinity faster than the RHS; and the LHS is a linear function, while the RHS is strictly increasing and strictly concave (first derivative strictly positive AND STRICTLY DECREASING everywhere), so by continuity there must be at least one solution but by concavity there can be no more than one solution.
Then since $x_1 = 2$ and obviously $A \ge 123$, $x_1 < A$. Now show by induction that $x_n < A$ for all $n$. (USE the fact that the RHS of the above equation is a strictly increasing function of $t$.)
OK, so the sequence is bounded between 123 and $A$ (except the first term, which we can disregard). Now use (or prove) the following observation: IF a sequence is bounded, then EITHER (a) it converges to a single limit $l$, OR (b) it has at least two convergent subsequences with distinct limits.
In our problem, any finite limit of any subsequence of the given sequence must satisfy the equation 
$$ y = 123 + \sqrt{4 + 5 \sin y + 6 \sqrt y}$$
Note that this equation is DIFFERENT from the above equation in $t$ and it will have a solution $l$ different from $A$. Prove that THIS equation also has a unique solution. So alternative (b) in the earlier paragraph is not possible, and we find the sequence converges, and specifically to $l$.
I don't know how to SOLVE this equation other than by trial and error (or using a graphing calculator or Wolfram Alpha) but perhaps that is not the true point of this problem.
