What is the meaning of limit of Fibonacci sequence? For general Fibonacci sequence with $F_1=F_2=1$,
It is known that limit of $\frac {F_{n+1}} {F_n} $ exists.
I am wondering what this limit implies and why it is important to compare these two sequences.
Thanks 
 A: The recurrence $F_{i+2}=F_i+F_{i+1}$ gives rise to a simple recurrence relation for the ratio of successive terms $\frac{F_{i+1}}{F_i}\in\Bbb R\cup\{\infty\}$, namely $\frac{F_{i+2}}{F_{i+1}}=\frac{F_i}{F_{i+1}}+\frac{F_{i+1}}{F_{i+1}}=(\frac{F_{i+1}}{F_i})^{-1}+1$, which is of order$~1$. The dynamics of the map $x\mapsto x^{-1}+1$ on $\Bbb R\cup\{\infty\}$ is quite simple: the map has two fixed points $\frac{1+\sqrt5}2$ and $\frac{1-\sqrt5}2$, of which the first is attractive, ad the second is repulsive. For any sequence satisfying the recursion and with initial ratio not exactly equal to the second fixed point, the ratio of successive terms will converge to the the first fixed point as $n\to\infty\,$; this is in particular the case for the Fibonacci sequence. 
A: One reason to be interested in $\frac{F_{n+1}}{F_n}$ and its limit, is that it is a rational number which gives a good approximation to the golden ratio $\phi$ (which is irrational). In fact, in a way, this is the best rational approximation to $\phi$. Understanding why leads to the fascinating topic of continued fractions. The continued fraction expansion of $\phi$ is 
$$
\phi  = 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1+\ldots}}}.
$$
If you truncate this fraction after $n-1$ divisions you get the $n$-th convergent. The first convergent of $\phi$ is $1$, the second one is $1 + \frac{1}{1} = 2$, the third is $1 + \frac{1}{1 + \frac{1}{1}} = \frac{3}{2}$, the fourth is $1 + \frac{1}{1 + \frac{1}{1+\frac{1}{1}}} = \frac{5}{3}$. 
Hopefully you see the pattern: using induction you can show that the $n$-th convergent is $\frac{F_{n+1}}{F_n}$. The neat thing is that a basic theorem about continued fractions shows that if $\frac{p}{q}$ is the $n$-th convergent of the continued fraction of a number $x$, then any other rational number $\frac{r}{s}$ with $s \le q$ is a worse approximation to $x$. I.e. for all $r$ and all $s \le q$, we have
$$
\left|x - \frac{p}{q}\right| \le \left|x - \frac{r}{s}\right|.
$$
So, not only $\frac{F_{n+1}}{F_n}$ is a rational sequence that converges to $\phi$, but in fact it converges as fast as possible, in a certain precise sense.
Additionally, as noted in a comment, because all terms in the continued fraction expansion of $\phi$ are $1$, $\phi$ is the hardest irrational to approximate. This is the essence of Hurwitz's inequality.
A: The existence of the limit reflects the fact that the Fibonacci sequence is essentially a geometric sequence (it is actually a linear combination of two geometric sequences but one of them dominates the other). See Wikipedia.
A: Let $f_n$ be the original Fibonacci sequence ($f_0=0,f_1=1$). Then any sequence $F$ defined as
$$F_1=a\\F_2=b\\F_{n+2}=F_{n+1}+F_n$$
can be written as
$$F_n=\frac{(3a-b)f_n-(a-b)L_n}2$$
where $L_n$ is $n$-th Lucas number. Now, knowing that Fibonacci sequence is recurrence equation, it can be solved like this:
$$x^2=x+1\implies x_{1,2}=\frac{1\pm\sqrt5}2$$
Now, we can use these solutions to construct the solution for our recurrent equation (se "how to solve recurrent equation" article on Wikipedia):
$$f_n=\frac{x_1^n-x_2^n}{x_1-x_2}$$
Knowing it, you will see that $F$ is always an exponential sequence and therefore limit of it's two consecutive elements always exists.
A: Denote the limit with $L$. Then:
$$\lim_{n\to\infty} \frac{F_{n+1}}{F_n}=L \Rightarrow  \lim_{n\to\infty} \frac{1}{\frac{F_{n+2}-F_{n+1}}{F_{n+1}}}=L \Rightarrow \frac{1}{L-1}=L \Rightarrow L=\frac{1+\sqrt{5}}{2}=\phi.$$
