calculus - Does calculating the limit of a given real-valued (recursive) sequence already imply its convergence?

Assuming we are given a real-valued recursive sequence $(a_n)_{n \in \mathbb{N}}$ by its starting point $a_1$ and its recursive function $a_{n+1} = \varphi(a_n)$

EDIT: Since this question caused some confusion or i failed to be clear i'd like to add the recursive sequence:

$$a_1 = \sqrt{2},\ a_{n+1} = \sqrt{2a_n}$$

In order to prove the convergence of $(a_n)_{n}$ i would try to prove boundedness and monotonicity (if that's the case). Then i would try to compute its limit.

However, a few days ago i've been asked whether it would not be sufficient enough to simply compute $$\lim_{n \to \infty} a_n$$

without even proving whether or not $(a_n)_n$ is convergent at all.

Ive been asked that if we basically could find an $a$ for which $$ \lim_{n \to \infty} a_n = a$$ holds
we could immediately conclude that the given sequence was convergent in the first place.

I was quite sceptical but i failed to provide a reasonable answer why it would certainly not sufficient enough. My first reaction was that it might be possible that $a$ might not be the limit but only an accumulation point. But i'm not sure if that's either the case or if i'm completely mistaken.

Can you possibly help me with an answer to that question? Can i simply compute $\lim_{n \to \infty} a_n = a$ prior to proving the existence of such an $a$ at all?

This is all restricted to $\mathbb{R}$ but any general answer is highly appreciated as well.

Thank you very much for any help.

Answer

Take a sequence $(a_n)$ defined by $a_0=0$ and $a_n=2a_{n-1}+1$. If you assume that the limit $L=\lim_{n\to\infty}(a_n)$ exists then by taking the limit of both sides you get $L=-1$. Does this answer your question?