What is the limit of $a_{n+1}=\sqrt{7-(-1)^na_n}$? 
Find the limit of the following recurrence relation:
  $$a_{n+1}=\sqrt{7-(-1)^na_n}, n\geq 0$$ with $a_0=0$.

I have thought that we can transform the relation to the following: $$a_{n+1}^2+(-1)^na_n=7$$ but I cannot take it further!!! Any help? (I also don't know how to prove the convergence of such a sequence)
Edit 1: I think that it is not a duplicate, because my relation has also the $(-1)^n$ part.
 A: Hint. Note that the subsequences $(a_{2n})$ and $(a_{2n+1})$ have different 
limits.
Also, 
$$
b_{n+1}=a_{2n+2}=\sqrt{7-a_{2n+1}}=\sqrt{7-\sqrt{7+a_{2n}}}=\sqrt{7-\sqrt{7+b_n}}
$$
Similarly
$$
c_{n+1}=a_{2n+1}=\sqrt{7+\sqrt{7-a_{2n-1}}}=\sqrt{7+\sqrt{7-c_{n-1}}}
$$
Clearly, $(b_{2n})$, $(c_{2n})$, $(b_{2n+1})$, $(c_{2n+1})$ are monotonic, bounded and hence convergent.
For $(b_{n})$, if $x$ is the limit, then
$$
x=\sqrt{7-\sqrt{7+x}}
$$
or $(7-x^2)^2=7-x$...
A: One can show that the limit does not exist by contradiction. Suppose $\lim_na_n$ exists. Then we must have that $x=:\lim_na_{2n}$ and $y:=\lim_na_{2n+1}$ both exists. We show that $x\neq y$ and thus have a contradiction. 
By the recursive relation, one has
$$
a_{2n+1}=\sqrt{7-\sqrt{7+a_{2n-1}}},\quad
a_{2n}=\sqrt{7+\sqrt{7-a_{2n-2}}}
$$
which implies by taking the limit and the continuity of square root that
$$
y=\sqrt{7-\sqrt{7+y}}\qquad(1)\\
x=\sqrt{7+\sqrt{7-x}}\qquad(2)
$$
One can check that solutions to (1) and (2) can never be the same:


