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.

$$a_{n+1}-\frac{\sqrt5-1}{2}=\frac{\frac{\sqrt5-1}{2}-a_n}{\sqrt{1-a_n}+\frac{\sqrt5-1}{2}}$$ and
$$a_{n+2}-a_n=\sqrt{1-\sqrt{1-a_n}}-a_n=\frac{a_n\sqrt{1-a_n}\left(\frac{\sqrt5-1}{2}-a_n\right)\left(\frac{\sqrt5+1}{2}+a_n\right)}{(\sqrt{1-\sqrt{1-a_n}}+a_n)((1+a_n)\sqrt{1-a_n}+1)}.$$
Now, let $a_1<\frac{\sqrt5-1}{2}.$
Thus, for any odd $n$ we have $a_n<\frac{\sqrt5-1}{2}$ and $a_{n+2}>a_n$,
which says that for odd $n$ there is $\lim\limits_{n\rightarrow+\infty}a_n$ and by your work
(we just need to solve the following equation: $a=\sqrt{1-\sqrt{1-a}}.$) it's equal to $\frac{\sqrt5-1}{2}.$
Now, $$a_2=\sqrt{1-a_1}>\sqrt{1-\frac{\sqrt5-1}{2}}=\frac{\sqrt5-1}{2},$$ which says that for even $n$ we have $a_n>\frac{\sqrt5-1}{2}$ and $a_{n+2}<a_n$,
which gives that for even $n$ there is $\lim\limits_{n\rightarrow+\infty}a_n.$
Let this limit is equal $b$.
Thus, since $$a_{n+2}=\sqrt{1-\sqrt{1-a_n}}$$ and $f(x)=\sqrt{x}$ is a continuous function, we obtain:
$$b=\sqrt{1-\sqrt{1-b}},$$ which gives $$b\in\left\{0,1,\frac{\sqrt5-1}{2}\right\}.$$
Since, $a_{n+2}<a_n$, we see that $b\neq1$ and since, $a_n>\frac{\sqrt5-1}{2},$ we see that $b\neq0$.
Thus, $b=\frac{\sqrt5-1}{2}.$
Can you end it now?
