Existence of limit question

  • #1
DeadOriginal
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1
Hi all,

I have a quick question about limits. This is something I should know but shame on me I forgot.

If a function is bounded both above and below but isn't monotonic and is not necessarily continuous at all points, how do I go about proving its limit exists?

In particular I am thinking about something along the lines of this function:
$$
\lim\limits_{n\rightarrow\infty}\frac{F_{n+1}}{F_{n}}
$$
where $$F_{n}$$ is the nth Fibonacci number. I know that the limit exists and I know what that limit is but say I didn't. Say I wanted to prove that it did exist. How do I do that without using the closed form expression for Fibonacci numbers? I know that it is bounded but it isn't monotonic.

This wasn't a homework question but a question that came to my mind after proving that this limit was equal to the golden ratio using the closed form expression of the Fibonacci numbers.
 

Answers and Replies

  • #2
eigenperson
160
21
If a function is bounded both above and below but isn't monotonic and is not necessarily continuous at all points, how do I go about proving its limit exists?
Your question is stated in terms of functions, but your example problem is stated in terms of sequences. I'll answer for sequences.

Obviously there isn't one single method that can work in general, because the limit of such a sequence doesn't always exist.

Here are a few options to try:

1. If the differences of a sequence are alternating in sign and have limit zero, then the sequence has a limit. This is the easiest method to apply in this case. It doesn't tell you what the limit is, but you should be able to figure it out using some other method (for example, it must satisfy L = 1 + 1/L).

2. "Squeeze theorem": If you can construct an upper bound and a lower bound for the sequence such that both the upper bound and lower bound have limit L, then the sequence has limit L.

3. Prove the sequence is Cauchy.

4. Guess the limit and prove that the sequence approaches it using the definition of limit.
 
  • #3
Office_Shredder
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You know that Fn+1 = Fn+Fn-1, so if the limit exists:

[tex] \lim_{n\to \infty} F_{n+1}/F_n = \lim_{n\to \infty} (F_n+F_{n-1})/F_n = \lim_{n\to \infty} 1 + F_{n-1}/F_n [/tex]

If [itex] \lim_{n\to \infty} F_{n+1}/F_n = \alpha[/itex] then the rightmost guy above is equal to [itex] 1 + 1/\alpha[/itex] so if the limit exists we get
[tex] \alpha = 1+1/\alpha[/tex]

From which you can solve for what the limit is equal to. Ironically calculating the limit is much more difficult than showing the limit exists. In this case if I remember correctly for
[tex] a_n = F_{n+1}/F_n[/tex]
the a2n form a monotonic sequence and the a2n+1 form a monotonic sequence. The argument above with some tweaking shows that the numbers they converge to are the same, and therefore the sequence [itex] a_n[/itex] converges.
 
  • #4
DeadOriginal
274
1
You know that Fn+1 = Fn+Fn-1, so if the limit exists:

[tex] \lim_{n\to \infty} F_{n+1}/F_n = \lim_{n\to \infty} (F_n+F_{n-1})/F_n = \lim_{n\to \infty} 1 + F_{n-1}/F_n [/tex]

If [itex] \lim_{n\to \infty} F_{n+1}/F_n = \alpha[/itex] then the rightmost guy above is equal to [itex] 1 + 1/\alpha[/itex] so if the limit exists we get
[tex] \alpha = 1+1/\alpha[/tex]

From which you can solve for what the limit is equal to. Ironically calculating the limit is much more difficult than showing the limit exists. In this case if I remember correctly for
[tex] a_n = F_{n+1}/F_n[/tex]
the a2n form a monotonic sequence and the a2n+1 form a monotonic sequence. The argument above with some tweaking shows that the numbers they converge to are the same, and therefore the sequence [itex] a_n[/itex] converges.

Yes that is exactly what I wanted to do. I have a proof that the limit is equal to the golden ratio using the closed form expression but I had to do a lot of footwork to show that the Fibonacci numbers could in fact be expressed as that closed form (I did this using induction).

I was also able to show the limit was equal to the golden ratio by using the assumption that it was equal to a. What I had forgotten was how to show functions/sequences converged if I only had boundedness. Thank you and eigenperson for reminding me.

I guess it was a bad idea to leave home without Spivak...
 
  • #5
mathwonk
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have you thought about sin(x) as x --> infinity?
 
  • #6
DeadOriginal
274
1
That is definitely an interesting one. I have always found limits of trig functions to be kind of funky. I know that the limit doesn't exist. The only proof I know of to prove that is the epsilon delta proof that derives a contradiction.
 
  • #7
pbuk
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1. If the differences of a sequence are alternating in sign and have limit zero, then the sequence has a limit.

Is that true? What about the sequence [tex]S_n = (S_{n-1}+\frac{4}{n} \text{ where n is even, } S_{n-1} - \frac{2}{n-1} \text{ where n is odd}), S_0 = S_1 = 0 [/tex]Note that this generates the series ## \frac{4}{2} - \frac{2}{2} + \frac{4}{4} - \frac{2}{4} + \frac{4}{6} - \frac{2}{6} \ldots ##
 
  • #8
eigenperson
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Yeah, I should have said "decreasing" instead of "has limit zero".
 
  • #9
pbuk
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