Proof of a limit for a recursively-defined sequence I have a sequence defined by
$$
a_1=1,\quad a_2=1,\quad a_n=\frac{1}{a_{n-1}}+\frac{1}{a_{n-2}}\text{ for } n\ge2.
$$
Now, if $\lim_{n\to \infty}a_n=g$ then $\lim_{n\to \infty}a_n=\lim_{n\to \infty}\Bigl(\frac{1}{a_{n-1}}+\frac{1}{a_{n-2}}\Bigr)=\frac{2}{g}$ so $g=\sqrt{2}$ or $g=-\sqrt{2}$ but $a_n&gt;0$ so $g=\sqrt{2}$.

Now, how do I prove that it has an actual  limit? Also, it can be proven that $1\le a_n\le2$, and it's not monotonic because $a_4 \gt a_5 \lt a_6$.
Also, it's not monotonic after some $N\in\mathbb N$
As you mention, it is straightforward to prove by induction that $1 \le a_n \le 2$, for all $n \ge 1$.

Let $L = \limsup_{n\to\infty} a_n$, $l = \liminf_{n\to\infty} a_n$, we have $1 \le l, L\le 2$. For any $\varepsilon&gt;0$, we know that $l - \varepsilon \le a_n \le L + \varepsilon$, if $n$ is sufficiently large. Thus, for $n$ large enough
$$ \frac{2}{L+\varepsilon} \le a_n \le \frac{2}{l-\varepsilon}. $$
Since $\varepsilon$ can be chosen to be arbitrarily small, we get
$$ L = \limsup_{n\to\infty} a_n \le \frac{2}{l}, \,\, l = \liminf_{n\to\infty} a_n \ge \frac{2}{L},$$
hence $L \cdot l =2$.

Now, we may choose a subsequence $\{n_k\}$ such that, as $k\to\infty$,
$$ a_{n_k + 1} \to L,\, a_{n_k} \to l_1,\, a_{n_k - 1} \to l_2,\, a_{n_k - 2} \to l_3, $$
for some $l_1, l_2, l_3 \in [l,L]$. By the definition of the sequence $a_n$, we have
$$ \frac{2}{l} = L = \frac{1}{l_1} + \frac{1}{l_2},\quad l_1 = \frac{1}{l_2} + \frac{1}{l_3}. $$
The left equality implies $l_1 = l_2 = l$, and then, using $\frac{2}{L} = l = l_1$, the right equality gives $l_2 = l_3 = L.$ Therefore, $l = L$, and since $L \cdot l = 2$, we find that the limit is $\sqrt{2}$.
To be able to do $$g=\lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{1}{a_{n-1}}+\frac{1}{a_{n-2}}=\frac{2}{g}$$ you should prove first that the sequence is bounded and some how it gets arbitrarily close to the candidate for limit, and therefore convergent. For example, let $c_n$ be a sequence of positive real numbers, if for all $n\in\mathbb{N}$ $c_n\leq M$ for some $M\geq 0$ and for all $N\in\mathbb{N}$ exists $n_0\in\mathbb{N}$ such that $c_n\geq c_{N}$ for all $n\geq n_0$, then $c_n$ is convergent, not necessarily to $M$. Only then you're able to manipulate limits as you wish to find the actual limit.

Otherwise you may end up with stuff like this: Take $b_1=1$ and $b_n=1+b_{n-1}^{2}$, if $\lim_{n\to\infty}b_n=g$ then $$g=\lim_{n\to\infty}b_n=\lim_{n\to\infty}1+b_{n-1}^{2}=1+g^{2}$$ so $g=1\pm i\sqrt{3}$ which clearly doesn't make sense.