Let $\lbrace F_n\rbrace_{n \in \mathbb{N_0}}$ be the Fibonacci sequence.
$F_{n+1}=F_{n-1}+F_{n-2}$ for $n \in \mathbb{N}$ with $n \geq 2$ and start values $F_0:=0$ and $F_1:=1$.
How to determine:
$\lim\limits_{n\to\infty}\frac{F_n}{F_{n+1}}$?
I used:
If $a=\lim\limits_{n\to\infty}\frac{F_n}{F_{n+1}}$ then
$a=\lim\limits_{n\to\infty}\frac{F_n}{F_{n+1}}=\lim\limits_{n\to\infty}\frac{F_n}{F_n+F_{n-1}}$
Here I don't know how to continue.
Assign a variable to the first limit, such as $x$.
$$\lim_{n\to\infty}\frac{F_n}{F_{n+1}} = x$$
By definition, $F_{n+1} = F_n+F_{n-1}$, so rewrite the limit.
$$\lim_{n\to\infty}\frac{F_n}{F_{n}+F_{n-1}} = x$$
$$\lim_{n\to\infty}\frac{F_{n}+F_{n-1}}{F_n} = \frac{1}{x}$$
$$1+\lim_{n\to\infty} \frac{F_{n-1}}{F_n} = \frac{1}{x}$$
If $\lim_\limits{n\to\infty}\frac{F_n}{F_{n+1}} = x$, then $\lim_\limits{n \to \infty} \frac{F_{n-1}}{F_n} = x$ as well.
$$1 + x =\frac{1}{x}$$
$$x+x^2 = 1 \implies x^2+x-1 = 0 \implies x = \frac{-1\pm\sqrt{5}}{2}$$
Now, it's clear that only $x = \frac{-1+\sqrt{5}}{2}$ applies here, so $x = \frac{1}{φ}$, or the reciprocal of the golden ratio. The interesting thing is that this does not apply specifically to the Fibonacci sequence, but for Fibonacci-like sequences in general.
HINT
Solve the recurrence analytically, the characteristic polynomial would be $$ x^2-x-1=0 $$ and its roots $r,R$ would produce $$ F_n = Ar^n + BR^n... $$
Using Binet's formula: $$\lim_{n\to\infty} \frac{F_n}{F_{n+1}}=\lim_{n\to\infty} \frac{\frac1{\sqrt{5}}\left(\varphi^n-\phi^n\right)}{\frac1{\sqrt{5}}\left(\varphi^{n+1}-\phi^{n+1}\right)}=\lim_{n\to\infty} \frac{\varphi^n\left(1-\frac{\phi^n}{\varphi^n}\right)}{\varphi^{n+1}\left(1-\frac{\phi^{n+1}}{\varphi^{n+1}}\right)}=\frac1{\varphi},$$ because: $$\lim_{n\to\infty} \left(\frac{\phi}{\varphi}\right)^n=0,$$ because: $$\left|\frac{\phi}{\varphi}\right|=\left|\frac{\frac{1-\sqrt{5}}{2}}{\frac{1+\sqrt{5}}{2}}\right|=\left|\frac{3-\sqrt{5}}{-2}\right|\approx 0.38<1.$$
