For a sequence $\{f_n\}_{n\geq 1}$ defined as... For a sequence $\{f_n\}_{n\geq 1}$ defined as:
$f_n= \frac{F_n}{F_{n+1}}$, where $F_n$ is the $n^{th}$ term in the Fibonacci series, starting with a $1$ rather than $0$.
I wish to find: $\lim_{n \to \infty} f_n$.
I have done it in two ways and I get two different answers for both.
Method 1:
$\lim_{n \to \infty} f_n= \lim_{n \to \infty} \frac{F_n}{F_{n+1}}=\lim_{n \to \infty} \frac{1}{1+\frac{F{n-1}}{F_n}}$.
Now as $n \to \infty$, $(n-1)\to \infty$. Thus $F_{n-1} \to F_n$. Which means $\frac{F_n}{F_{n-1}}\to 1$.
Thus, $\lim_{n \to \infty} f_n=\frac{1}{1+1}=\frac{1}{2}$.
But this method must be wrong because as per this logic as $n\to \infty$, $(n+1)\to \infty$. Thus, $\frac{F_n}{F_{n+1}}\to 1$, which means $f_n\to 1$. But $1\neq \frac{1}{2}$. Thus I would end up with a value that contradicts my previous value. Now, I know that it is wrong because I received a contradictory value. Can someone explain theoretically why this method is wrong?
Method 2:
$\lim_{n \to \infty} f_n= \lim_{n \to \infty} \frac{F_n}{F_{n+1}}=\lim_{n \to \infty} \frac{1}{1+\frac{F{n-1}}{F_n}}$.
Now as $n \to \infty$, let $\frac{F_n}{F_{n+1}} \to L$. Thus, $\frac{F{n-1}}{F_n} \to L$.
Thus we get $L=\frac{1}{1+L}$, which on solving we get $L=\frac{\sqrt{5} -1}{2}$. Is this approach correct?
If my second method is also wrong, please suggest and alternative method.  
 A: Let we prove something slightly more general. Assuming that the sequence $\{a_n\}_{n\geq 1}$ fulfills
$$ a_1=a_2=1,\qquad a_{n+2} = K a_{n+1} + a_n,\qquad K>0, \tag{1} $$
the characteristic polynomial is $p(x)=x^2-Kx-1$. Since the discriminant of $p(x)$ is $K^2+4>0$, $p$ has two distinct real roots, whose product is minus one by Viète's theorem. Let $\xi$ the root with absolute value $>1$ and $\eta$ the root with absolute value $<1$. We have an explicit formula:
$$ a_n = A \xi^n + B\eta^n \tag{2}$$
that, assuming $A\neq 0$, ensures:
$$ \lim_{n\to +\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to +\infty}\frac{A \xi^{n+1}+o(1)}{A\xi^n +o(1)} = \xi.\tag{3}$$
In our case $\left(K=1,\; A=\frac{5+\sqrt{5}}{10}\right)$ the last identity gives:

$$ \lim_{n\to +\infty}\frac{F_n}{F_{n+1}}=\frac{1}{\xi}=-\eta=\color{red}{\frac{\sqrt{5}-1}{2}}.\tag{4}$$

A: The first is clearly wrong, because it is based on the meaningless statement $F_{n-1}\to F_n$. Actually, the 'corrected' formulation: $F_{n-1}/F_n\to 1$ would be meaningful, but is it wrong. 
The second argument is based on the hypothesis $ F_{n-1}/F_n\to L$ which is correct for a suitable $L$. However this should be proved. The next stage of the argument, that is $L=1/(L+1)$ is correct. I make a little more explicit the argument. Starting from the definition of Fibonacci sequence,
$$
F_{n+2}=F_{n+1}+F_n
$$ 
one obtains after division (all terms are positive)
$$
1=\frac{F_{n+1}}{F_{n+2}}+\frac{F_{n}}{F_{n+2}}.
$$
If you know that $ F_{n}/F_{n+1}\to L$, then $\frac{F_{n}}{F_{n+2}}\to L^2$ and
$$
1=L+L^2
$$
the correct equation you obtained in the second argument.
Of course this is based on the assumption that $ F_{n-1}/F_n\to L$. This is obvious if you know the explicit power-like formula for $F_n$. But I think that it is not the spirit of the exercise.
A: We can try the following: using the well known symbol for the goldent ration
$$\phi:=\frac{1+\sqrt5}2.\;\;\text{Observe that}\;\;\frac1\phi=\frac2{1+\sqrt5}=\frac{2-2\sqrt5}{-4}=\frac{-1+\sqrt5}2<1$$
We also have
$$f_n=\frac{F_n}{F_{n+1}}=\frac{F_{n+1}-F_{n-1}}{F_{n+1}}=1-\frac{F_{n-1}}{F_{n+1}}\implies$$
$$\left|f_n-\left(1+\frac1\phi\right)\right|=\left|\frac{F_{n-1}}{F_{n+1}}+\frac1\phi\right|=\left|\frac{\phi F_{n-1}+F_{n+1}}{\phi F_{n+1}}\right|\le$$
$$\le \left(\frac1\phi\right)\left|F_{n+1}+\phi F_{n-1}\right|\le\ldots\le \left(\frac1\phi\right)^{n-1}\left|F_2+\phi F_0\right|\xrightarrow[n\to\infty]{}0$$
since $\;\phi^{-1}<1\;$ , and from here
$$\lim_{n\to\infty}f_n=1+\frac1\phi=\phi$$
