Why can you place in the recursive definition to find the limit? When required to find limits of recursive sequences, i.e.
$$x_{n+1}=\frac{1}{4-x_n}\qquad x_0=3$$
The steps are usually pretty consistent. First you prove it's monotonous and bounded, and therefore converges, then you place $L$ in the equation and compute the limit itself.
$$L=\frac{1}{4-L}\implies L=1\pm \frac{\sqrt{3}}{2}$$
And then you choose the lower one because it's the real limit.
Now the placement seems to be a recurring motive, but I just can't understand why it's true, especially since you get 2 "limits" where one of them is just magically right.
The only explanation I seem to find is that $x_{n+1}$ and $x_n$  are actually the same sequence, which I sort of intuitively understand, but I'm still uncertain of. Is there any rigorous justification for this placement?
 A: The reason we can replace $x_n$ and $x_{n+1}$ with the same $L$ is because they have the same limit. To see why this is true we need the definition of the limit of a sequence: we say $x_n \to L$ when for all $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that for all $n\ge N$ we have
$$ |x_n - L| < \epsilon.$$
Since you have proved that $x_n$ converges we can use this definition to show that $x_{n+1}$ also converges, and to the same limit. Let $\epsilon > 0$. Let's say $x_n$ converges to some real number $L$. Then by the definition of the limit we can choose $N\in \mathbb{N}$ such that if $n\ge N$ then $|x_n - L| < \epsilon$. Now given $n\ge N$ we have $n+1 > n \ge N$ so that
$$
|x_{n+1} - L| < \epsilon.
$$
Thus, by the definition of the limit, $x_{n+1} \to L$. 
Now that we have established that $x_{n+1}$ and $x_n$ have the same limit $L$ we can compute the limit of the recursion. Taking the limit of both sides,
$$
\lim_{n\to\infty}x_{n+1} = \lim_{n\to\infty} \frac{1}{4-x_n}.
$$
Now using that $x_n \to L$ and $x_{n+1} \to L$,
$$
L = \frac{1}{4-L}.
$$
We can expand and simpify this to get the quadratic 
$$
L^2 - 4L + 1 = 0
$$
which, by the quadratic formula, has roots 
$$
L = 2 \pm \sqrt{3}.
$$
Now the reason why $L$ is "magically" $2- \sqrt{3}$ is because it's the only root that makes sense: note that $x_n$ is monotone decreasing and bounded below by $2 - \sqrt{3}$ so it converges to $2-\sqrt{3}$ by the monotone convergence theorem.
