<s>[INST] How to find $\lim\limits_{n\to +\infty}(3+\frac{1}{a_n})$, where $a_n = 3+\frac{1}{a_{n-1}}$? How to find $\lim\limits_{n\to +\infty}(3+\frac{1}{a_n})$, where $a_n = 3+\frac{1}{a_{n-1}}$?
I tried to find $(a_n-a_{n-1})$ but it didn't help.
 [/INST] Let's show that for $\forall a_0>0$, the sequence is converging ...
Proposition 1. $3< a_n < 4$, for $\forall n \geq 2$.
If $a_0 > 0 \Rightarrow \frac{1}{a_0}>0 \Rightarrow \color{red}{a_1}=3+\frac{1}{a_0}\color{red}{>3}$. So $a_1 >0 \Rightarrow \frac{1}{a_1}>0 \Rightarrow \color{red}{a_2}=3+\frac{1}{a_1}\color{red}{>3}$ and, by induction 
$$\color{red}{a_n > 3}, \forall n >0 \tag{1}$$
Also, from $(1)$ $a_n > 3 \Rightarrow 0< \frac{1}{a_n} < \frac{1}{3} < 1 \Rightarrow \color{red}{a_{n+1}}=3+\frac{1}{a_n}\color{red}{<4}$. But $(1)$ is true for $n > 0$, so (by induction again) 
$$\color{red}{a_n < 4}, \forall n>1 \tag{2}$$
Altogether $$3< a_n < 4,\forall n \geq 2$$

Now, let's look at the function 
$$f(x)=3+\frac{1}{x}$$
which is an important one, because $a_{n+1}=f(a_n)$.
Proposition 2. if $3\leq x \leq 4 \Rightarrow 3 \leq f(x) \leq 4$
$3\leq x \leq 4 \Rightarrow \frac{1}{3} \geq \frac{1}{x} \geq \frac{1}{4} \Rightarrow 4 > 3+ \frac{1}{3} \geq 3+\frac{1}{x} \geq 3+\frac{1}{3} > 3$ which is $3 < f(x) < 4$

Proposition 3 Function $f: [3,4] \rightarrow [3,4]$ is a contraction mapping.
From MVT 
$$|f(x)-f(y)|= |f'(\varepsilon)| |x-y|=\left|\frac{1}{\varepsilon^2}\right| |x-y|$$
where $\varepsilon \in (x,y) \subset [3,4]$, thus
$$|f(x)-f(y)| \leq \frac{1}{9} |x-y|$$

Altogether, we can ignore the first 2 elements (Proposition 1) of the sequence, when the initial element is greater than $0$, they don't affect the convergence of the sequence. And according to the Banach fixed-point theorem, the sequence has a limit. More reading on this subject here
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