If the sequence Xn tends to a limit L then
L = 1
1+ L
 .
Hence L2+L-1=0 and , solving this quadratic equation and taking the positive solution as L cannot be negative, L=(Ö5 - 1)/2.

The solutions of the quadratic equation x2 - x - 1 = 0 are x=(1 ±Ö5)/2 and the Golden Ratio f, which is positive, is (1 + Ö5)/2. Notice that L = f- 1 and Lf = (Ö5 - 1)(Ö5 + 1)/4 = 1 so
L = 1
f

.

We next prove that
Xn= Fn
Fn+1
 .
This is true for n=1 and n=2:
X1 = F1
F2
 = 1
1
 = 1,     X2= F2
F3
 = 1
2
 .
If the theorem is true for Xk then
Xk+1= 1
1 + Xk
 = 1
1+ \strut Fk
Fk+1
 = Fk+1
Fk+1+Fk
 = Fk+1
Fk+2
 ,
and hence it is rue for n=k+1 so by the axiom of induction it is true for all n.

Hence the ratio of successive terms of the Fibonacci sequence
Fn+1
Fn
 = 1
Xn
and as
Xn ® 1
f

as n ® ¥ this ratio tends to the Golden Ratio as n® ¥.