The Fibonacci numbers $x_1,x_2,.......,$ are defined recursively by $x_1=1, x_2=2$ and $x_{n+1}=x_n+x_{n-1}$ for $n\geq2$. Show that, $\lim_{n \to \infty}\frac{x_{n+1}}{x_n}$ exists, and evaluate the limit.
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Try using the matrix representation and the spectral theorem. This is going to be fun. – Thomas Aug 31 '15 at 07:19
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Are you trying to do this without first establishing the formula for $x_n$? – 2'5 9'2 Aug 31 '15 at 07:21
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This question is asked here – GAVD Aug 31 '15 at 10:38
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Define $a_n = \frac{x_{n+1}}{x_n}$
Then, you can show that $$a_{n+1} = \frac{x_{n+2}}{x_{n+1}} = \frac{x_{n+1} + x_n}{x_{n+1}} = 1 + \frac{x_n}{x_{n+1}} = 1+\frac1{a_n}$$
So you now have a function $f$ such that $a_{n+1} = f(a_n)$.
Now you can use the standard methods for recursively defined sequences to prove that the limit $a$ exists and that if it exists, it satisfies the equation $a=1+\frac1a$
5xum
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you should work out some details so OP can understand. rushing to post a solution is "counter productive" here...OP is not at your level... – DeepSea Aug 31 '15 at 07:30
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2@Ganymede I don't know the level of the OP, and neither do you. If my answer is above his level, he can ask me about it in the comment. I assumed he already knows how to deal with limits of the type $a_n=f(a_{n-1})$ – 5xum Aug 31 '15 at 07:32