When I started with this question, I wanted to know why my reasoning was wrong. Nevertheless, after checking some examples, I've noticed that my conjecture was actually - or at least seems to be - right! Therefore I'm looking for counterexamples or rigorous proofs.
Suppose that a recursive sequence with $L_0=5$ is defined recursively as follows
$$L_{n+1}=\frac{L_n+1}{L_n+2}$$
Sidenote: Please notice that this is only an example.
When it comes to evaluating the limit of a recursive sequence, you're trying to find $$\lim_{n\to\infty} \bigg(\frac{L_n+1}{L_n+2}\bigg)=\lim_{n\to \infty}L_{n+1}\color{red}{\stackrel{?}{=}}\lim_{n\to \infty}L_{n}$$
At least to me, it sounds intuitive that, if the sequence has an upper bound, and the difference between consecutive terms approaches 0 $$L_n\sim L_{n+1}\quad\text{as $n$ approaches infinity}$$
Therefore, evaluating an upper bound should be a simple as solving the equation $$L_n=L_{n+1}\stackrel{\text{in the example}}{\implies}L_n=\frac{L_n+1}{L_n+2}\implies L_n=\frac{-1+\sqrt5}{2}$$
Another example would be, for instance, the sequence defined recursively as follows $$M_{n+1}=\frac{M_n}{3}+\frac{1}{M_n}\qquad \text{where}\quad M_0=4$$ which converges to $\sqrt{\frac{3}{2}}$.
This leads to
Conjecture
Suppose there's a recursive sequence $\{L_n\}^\infty_{n=0}$ that converges to $k$, where $L_{n+1}-L_n\to 0$ as $n$ approaches infinity. Then, $k$ is one of the solutions to the equation $$L_n=L_{n+1}$$
If $\lim_{n \to \infty} L_n = L$, then this is also $\lim_{n \to \infty} L_{n+1}$ (or, for that matter, $L_{n+k}$ for any $k$). So if the sequence satisfies some equation $f(L_n, L_{n+1}) = 0$ where $f$ is a continuous function on some subset of $\mathbb R^2$ containing $(L_n, L_{n+1})$ for all $n$ and also containing $(L,L)$, we have $$ f(L,L) = \lim_{n \to \infty} f(L_n, L_{n+1}) = 0$$