Find limit of $a_{n+1} = \frac{2a_n}{a_n+1}$. 
Determine if the following sequence $(a_n)_{n \in \mathbb{N}}$ converges and if so, find the limit. $a_1 \geq 0$, $a_{n+1} = \frac{2a_n}{a_n+1}$.

What I've done so far:
First we can notice that if $a_1 = 0 \Rightarrow a_n = 0, n \in \mathbb{N}$ and thus $\lim\limits_{n \to \infty} a_n=0$.
If $a_1 > 0, a_n > 0$ by induction and $a_{n+1} = 2 - \frac{2}{a_n + 1}$ which gives $a_{n+1}a_n + a_{n+1}-2a_n = 0$. I'm not familiar with recurrence equations, but WolframAlpha gives an equation such that a limit of it is easy to evaluate and equals $1, n \rightarrow \infty$.
Is there another way to find the limit of this sequence?

Let $f(x) = \frac{2x}{x+1}$ for $x\geq 0$, so that $a_{n+1}= f(a_n)$.
Then, we have :
$$\forall x\geq 0, f(x)-x = \frac{x(1-x)}{x+1}$$
Therefore $1>f(x) > x$ when $x\in (0,1)$ and $1<f(x)<x$ when $x>1$.

*

*If $a_1$ is equal to $0$ or $1$, then $f(a_0) = a_0$ and the sequence is constant.


*If $a_1 \in (0,1)$, then you can show by induction that for all $n\in \mathbb N$,
$$0<a_n<a_{n+1}<1$$
Therefore, $(a_n)$ converges to some $\ell \in (0,1]$. Taking the limit in $a_{n+1} = f(a_n)$, we find $\ell = f(\ell)$, whose only solution on $(0,1]$ is $\ell = 1$.


*If $a_1 \in (1,+\infty)$, then by induction :
$$\forall n\in\mathbb N, 1<a_{n+1}<a_n $$
Therefore $a_n \to \ell \geq 1$. Again we have $f(\ell) = \ell$ which implies $\ell =1$.
Conclusion

If $a_1 = 0$, then $(a_n)$ converges to $0$. If $a_1>0$, then $(a_n)$ converges to $1$.

