Find Limit of Fibonacci Sequence | Determine L

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Let f[itex]_{n}[/itex] be the Fibonacci sequence and let [itex]x_{n}[/itex] = [itex]f_{n+1}[/itex]/[itex]f_{n}[/itex]. Given that lim[itex](x_{n}[/itex])=L exist determine L.

Ok so I know that the limit is [itex]\frac{1+\sqrt{5}}{2}[/itex] from previous experience with the sequence, but I am not sure how do you show that without writing out a lot of terms and then noticing what I all ready know it is. How do you find the limit of a sequence to a number if your not given any numbers to work with?
 
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true, but how do you ever get a number when you are dealing with f[itex]_{n}[/itex] and f[itex]_{n+1}[/itex]. How can you just make a jump and say there is a [itex]\sqrt{5}[/itex] in there
 
Ok I am not sure how you got there but this is what I have so farlet [itex]x_{n}[/itex] = [itex]f_{n+1}[/itex]/[itex]f_{n}[/itex] and let lim[itex](x_{n}[/itex])=L. From here is where I get stuck. I know that every sub sequence of a convergent sequence converges to the same number by some theorem, but I am not sure how that is at all helpful.

Would you do something like
[itex]x_{n}[/itex] = [itex]f_{n+1}[/itex]/[itex]f_{n}[/itex]
[itex]x_{n-1}[/itex] = [itex]f_{n}[/itex]/[itex]f_{n-1}[/itex]
 
Ok so you get

= [itex]f_{n}[/itex]/[itex]f_{n-1}[/itex] = [itex]f_{n+1}[/itex]/[itex]f_{n}[/itex]

I think I see where you got the quadratic equation now

[itex]f_{n}[/itex] ^2= [itex]f_{n-1}[/itex] [itex]f_{n+1}[/itex]=[itex]f_{n}[/itex] ^2 - [itex]f_{n-1}([/itex] [itex]f_{n+1}[/itex])to use the quadratic equation is this [itex]f_{n-1}([/itex] [itex]f_{n+1}[/itex]) considered b or c?

and that doesn't look very pretty to solve...
 
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Ok

[itex]f_{n+1}[/itex]=[itex]\frac{f_{n}}{f_{n-1}}[/itex]


[itex]f_{n}[/itex] ^2 - f[itex]_{n-1}[/itex]* [itex]\frac{f_{n}}{f_{n-1}}[/itex]

[itex]f_{n}[/itex] ^2 -[itex]f_{n}[/itex]=0

is that right?
 
oh so
f[itex]_{n+1}[/itex]= f[itex]_{n}[/itex] +f[itex]_{n-1}[/itex]


do I use that for f[itex]_{n+1}[/itex]
Also how do you know when to use what?
 
Punkyc7 said:
do I use that for f[itex]_{n+1}[/itex]

yes … that should give you your quadratic equation :wink:
Also how do you know when to use what?

You're told it's a Fibonacci sequence, so you obviously have to use that information somewhere! :smile:

and now I'm off to bed :zzz:
 
let [itex]x_{n}[/itex] = [itex]f_{n+1}[/itex]/[itex]f_{n}[/itex] and let lim[itex](x_{n}[/itex])=L. Since we know the sequence converges we can say

[itex]x_{n}[/itex] =[itex]x_{n-1}[/itex] Which Implies

= [itex]f_{n}[/itex]/[itex]f_{n-1}[/itex] = [itex]f_{n+1}[/itex]/[itex]f_{n}[/itex]

[itex]f_{n}[/itex] ^2= [itex]f_{n-1}[/itex] [itex]f_{n+1}[/itex]=[itex]f_{n}[/itex] ^2 - [itex]f_{n-1}([/itex] [itex]f_{n+1}[/itex])=0

[itex]f_{n}[/itex] ^2 - f[itex]_{n}[/itex]f[itex]_{n-1}[/itex]-f[itex]_{n-1}[/itex]^2=0How do you hammer this into the quadratic equation I am thinking the a=1 b=not sure c=not sure ? Also how do you get numbers from this when we don't have a single number?
 
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Punkyc7 said:
let [itex]x_{n}[/itex] = [itex]f_{n+1}[/itex]/[itex]f_{n}[/itex] and let lim[itex](x_{n}[/itex])=L. Since we know the sequence converges we can say

[itex]x_{n}[/itex] =[itex]x_{n-1}[/itex] Which Implies

= [itex]f_{n}[/itex]/[itex]f_{n-1}[/itex] = [itex]f_{n+1}[/itex]/[itex]f_{n}[/itex]
...which also equals L:

[itex]\frac{f_n}{f_{n-1}} = \frac{f_{n+1}}{f_n} = L[/itex]

Now take this portion:
[itex]\frac{f_{n+1}}{f_n} = L[/itex]

Replace the numerator with its equivalent, and then rewrite as a sum of two fractions. A substitution can be made, and you will end up with an expression on the left side with NO f's. Soon you will see a quadratic equation in terms of L. Solve for L.