Prove that $\lim_{n\to \infty} a_n=\frac{\sqrt{5}-1}{2}$ if $a_{n+1}=\sqrt{1-a_n}$ and $0If $0<a_0<1$ and $a_{n+1}=\sqrt{1-a_n}$, prove that:
$$\lim_{n\to \infty} a_n=\dfrac{\sqrt{5}-1}{2}$$
Here what I do is that when $n\to \infty$, $a_{n+1}=a_n$
$\therefore a_n=\sqrt{1-a_n}\Rightarrow a_{n}^2+a_n-1=0$
which implies that when $n\to \infty$, $a_n\to \dfrac{\sqrt{5}-1}{2}$
$\therefore \lim_{n\to \infty} a_n=\dfrac{\sqrt{5}-1}{2}$
My doubt here is, was I correct in simply assuming that $a_{n+1}=a_n$ when $n\to \infty$? (I'm sorry for I may be weak in basics.)
If not, then please suggest alternative method.

hint
It is easy to prove by induction that for all $ n\ge 0$, $0\le a_n\le 1$.
Let $$f(x)=\sqrt{1-x}$$
from $ [0,1)] $ to $[0,1)$.
$$f'(x)=\frac{-1}{2\sqrt{1-x}}<0$$
$ f$ is decreasing at $ [0,1) $, so the subsequences $ (a_{2n})$ and $( a_{2n+1})$ are monotonic and convergent.
I let you to prove they have the same limit.
