Limit of $a_{n}^{2}$ when the sequence is recursive The problem states:
Given the sequence:
$$a_0 \in \left(0,1\right)$$
$$a_{n+1}= 1 - \sqrt {1-a_{n}^{2}}$$
First part. 
Check whether it converges, and if so find its limit.
Second part. 
Find:
$$\lim_{n \to \infty} \frac {a_{n}^{2}}{a_{n+1}}$$
The first part is clear to me, and I've done it. I've proven that the sequence is bounded (0 from below and 1 from above), and also that it is monotone decreasing, so the sequence converges to 0.
The second part is what confuses me. If the sequence is given recursively, how do I know what $a_n$ even is? Furthermore, how do I put it in a limit that I'm supposed to evaluate?
Any ideas? 
Thanks.
 A: I would like to point out that the fact that $a_n$ is bounded below by $0$ and decreasing only implies that it converges and that its limit $L$ satisfies $L\geqslant 0$.
In fact, in this case $L$ must be $0$, but this must be deduced, for instance via
\begin{align}
L = 1 - \sqrt{1-L^2}
&\implies (L-1) = -\sqrt{1-L^2}
\\
&\implies (L-1)^2 = 1-L^2
\\
&\implies 2L^2-2L = 0
\\&\implies L(L-1) = 0 \implies L=0 \text{ or } L=1.
\end{align}

As for the second part, consider that your recurrence relation implies
\begin{align}
&(a_{n+1}-1) = -\sqrt{1-a_n^2}
\\
\implies &(a_{n+1}-1)^2 = 1-a_n^2
\\
\implies &a_n^2 = 1 - (a_{n+1}^2-2a_{n+1}+1) = 2a_{n+1} - a_{n+1}^2.
\end{align}
It follows that
\begin{align}
\lim_{n\to\infty}\frac{a_n^2}{a_{n+1}}
&=
\lim_{n\to\infty}\frac{2a_{n+1} - a_{n+1}^2}{a_{n+1}}
\\&=
\lim_{n\to\infty} 2 - a_{n+1} = 2,
\end{align}
where the last step follows from your previous calculation.
A: Since $a_n \to 0$ we have
$$\frac{a_n^2}{a_{n+1}}=\frac{a_n^2}{1-\sqrt{1-a_n^2}}=\frac{a_n^2}{1-\sqrt{1-a_n^2}}\,\frac{1+\sqrt{1-a_n^2}}{1+\sqrt{1-a_n^2}}=$$$$=\frac{a_n^2\left(1+\sqrt{1-a_n^2}\right)}{1-1+a_n^2}=1+\sqrt{1-a_n^2}\to 2$$
