Find the limit of $a_1 = 1 $ , $ a_{n+1} = \frac{\sqrt{1+a_n^2}-1}{a_n}$ , $n \in \mathbb{N}$ $a_1 = 1 $ ,  $ a_{n+1} = \frac{\sqrt{1+a_n^2}-1}{a_n}$ , $n \in \mathbb{N}$
It is easy to prove that the limit exists: the boundary of that expression is $0$ and it is monotonically decreasing.
The problem is to actually find the limit (it is $0$) because if I take arithmetic of limits:
$ \lim_{x \to +\infty} a_{n+1} =  \frac{\sqrt{1+ \lim_{n \to +\infty} a_n^2+1}}{\lim_{n \to +\infty} a_n}$
$g = \frac{\sqrt{1+g^2+1}}{g}  \implies g^2 = {\sqrt{1+g^2}-1} \implies g^2 + 1 = {\sqrt{1+g^2}} \implies 0 = 0$

$$a_{n+1} = \frac{\sqrt{1+a_n^2}-1}{a_n}=\\ \frac{\sqrt{1+a_n^2}-1}{a_n}.\frac{\sqrt{1+a_n^2}+1}{\sqrt{1+a_n^2}+1}=\\\frac{1+a_n^2-1}{a_n(\sqrt{1+a_n^2}+1)}=\\
\frac{a_n}{(\sqrt{1+a_n^2}+1)}\leq \frac{a_n}{(\sqrt{1}+1)}=\frac{a_n}{2} $$so $$0<a_{n+1}\leq \frac 12a_n\leq\frac14 a_{n-1}\leq\frac18a_{n-2}\cdots<1$$
