Sequel, recursion. limit

I have the following question:

For arbitrary $a_0\in\mathbb{R}$, $0<a_0<1$, is $a_{n+1}:=1-\sqrt{1-a_n}$ a real sequel $a_n$.

I have to show that:

$\lim_{n\to\infty} \frac{a_{n+1}}{a_n}=\frac{1}{2}$

I proofed befor, that the

$\lim_{n\to\infty} a_n=0$

I do not know how to start. Do you have any tips?

Hint: $$\frac{a_{n+1}}{a_n} = \frac{a_{n+1}\left(1+\sqrt{1-a_n}\right)}{a_n\left(1+\sqrt{1-a_n}\right)} =\frac{1}{1+\sqrt{1-a_n}}.$$
• I thought about using $(a+b)(a-b)=a^2-b^2$ – user105916 Nov 6 '13 at 16:51
• But than there is the a_n left in the numerator. How do you get the 1 and this denominator? But the hint is clear I think, because now follows $\lim_{n\to\infty}\frac{1}{1+\sqrt{1-a_n}}=\frac{1}{1+\sqrt{1-0}}=\frac{1}{2}$ – user105916 Nov 6 '13 at 16:54