evaluate limit of a sequence The problem is:
Prove the convergence of the sequence 
$\sqrt7,\; \sqrt{7-\sqrt7}, \; \sqrt{7-\sqrt{7+\sqrt7}},\; \sqrt{7-\sqrt{7+\sqrt{7-\sqrt7}}}$, ....
AND evaluate its limit.
If the convergen is proved, I can evaluate the limit by the recurrence relation
$a_{n+2} = \sqrt{7-\sqrt{7+a_n}}$.
A quickly find solution to this quartic equation is 2; and other roots (if I find them all) can be disposed (since they are either too large or negative).
But this method presupposes that I can find all roots of a quartic equation.
Can I have other method that bypasses this?
For example can I find another recurrence relation such that I dont have to solve a quartic (or cubic) equation? or at least a quintic equation that involvs only quadratic terms (thus can be reduced to quadratic equation)?
If these attempts are futile, I shall happily take my above mathod as an answer.
 A: We have $a_1=\sqrt 7$, $a_2=\sqrt{7-\sqrt 7}$, and then the recursion $a_{n+1}=f(a_n):=\sqrt{7-\sqrt{7+a_n}}$.
By induction, one quickly shows $0<a_n\le \sqrt 7$.
For $0\le x<y\le\sqrt 7$, we have $$0<\sqrt{7+y}-\sqrt {7+x}=\frac{y-x}{\sqrt{7+x}+\sqrt{7+x}}<\frac{y-x}{2\sqrt 7} $$
and for $0\le x<y<6$, $$0<\sqrt{7-x}-\sqrt{7-y}=\frac{y-x}{\sqrt{7-x}+\sqrt{7-y}} <\frac{y-x}2.$$
We conclude that for $x,y\in[0,\sqrt 7]$, also $f(x),f(y)\in[0,\sqrt 7]$ and $|f(x)-f(y)|\le \frac1{|x-y|}{4\sqrt 7}$, i.e., $f$ is a contraction map.
Therefore the even and the odd subsequence both convereg to a fixpoint of $f$ in $[0,\sqrt 7]$.
Remains to show that $f$ has exactly one fixpoint $a$ in that interval.
From $f(a)=a$, we get
$$ (7-a^2)^2-7=a,$$
or
$$\tag1a^4-14a^2-a+42=0.$$
The derivative of this, $4a^3-28a-1=4a(a^2-7)-1$, is $\le -1$ for  $a\in[0,\sqrt 7]$, hence at most one solution to $(1)$ can exist there.
With the rational root theorem in mind or by pure luck, we find that $a=2$ is a and hence the solution.
A: To prove the limit exists, show that$$a_{4n}>a_{4n+1}>2>a_{4n+3}>a_{4n+2}$$Using induction. For example,$$a_{4n}>2\implies\underbrace{a_{4n+2}=\sqrt{7-\sqrt{7+a_{4n}}}<\sqrt{7-\sqrt{7+2}}=2}_{\huge a_{4n+2}<2}$$Same with
$a_{4n+1}\implies a_{4n+3},\\a_{4n+2}\implies a_{4n+4},\\a_{4n+3}\implies a_{4n+5}.$
Likewise, don't forget to check $a_0$ and $a_1$. And then show that$$a_{4n}>a_{4n+1}>a_{4n+4}>a_{4n+5}\\a_{4n+7}>a_{4n+6}>a_{4n+3}>a_{4n+2}$$So that we can see that $a_n$ is bounded between $a_0$ and $a_2$, and the subsequences $a_{4n}$ and $a_{4n+1}$ are monotone decreasing and $a_{4n+2}$ and $a_{4n+3}$ are monotone increasing. From there, it simply involves showing that they must converge to $2$.
