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}$
$g = \frac{\sqrt{1+g^2+1}}{g}  \implies g^2 = {\sqrt{1+g^2}-1} \implies g^2 + 1 = {\sqrt{1+g^2}} \implies 0 = 0$

You can prove it directly like this
$$a_{n+1} = \frac{\sqrt{1+a_n^2}-1}{a_n} < \frac{\sqrt{1+a_n^2+\frac{1}{4}a_n^4}-1}{a_n} = \frac{1+\frac{1}{2}a_n^2-1}{a_n} = \frac 12 a_n$$
Therefore $a_n \to 0$, $n\to \infty$.
