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pivoxa15

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How do you prove the existence but not necessarily the value of a limit?

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In summary: Okay, how about a_n-a_{n-1} \le k \left( \frac{1}{n}-\frac{1}{n-1} \right) for all n greater then M?That's a pretty interesting example. Can you find a similar example on the reals?Take the real part of it and see what happens.

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pivoxa15

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How do you prove the existence but not necessarily the value of a limit?

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- #2

John Creighto

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pivoxa15 said:How do you prove the existence but not necessarily the value of a limit?

Why not give an example of such a problem because I think your question is too general.

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pivoxa15

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It is a general question.

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John Creighto

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pivoxa15 said:

It is a general question.

Which means there is about a million answers. For instance:

http://en.wikipedia.org/wiki/Absolute_convergence

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pivoxa15

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John Creighto

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pivoxa15 said:

Well how about showing the limit of [tex]a_n-a_{n+1}=0[/tex] as n->oo

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morphism

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This doesn't imply that (a_n) will converge. [Counterexample: a_n = log(n).]John Creighto said:Well how about showing the limit of [tex]a_n-a_{n+1}=0[/tex] as n->oo

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morphism

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If you're going to be taking a limit of something, then it had better be a sequence (or something sequence-like). So, no, you haven't narrowed down anything. Do you mean sequences of numbers?pivoxa15 said:

In any case, there are a few existence results that can come in handy. One example is the monotone convergence theorem, which states that if a sequences of real numbers is monotone and bounded, then it must converge.

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John Creighto

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morphism said:This doesn't imply that (a_n) will converge. [Counterexample: a_n = log(n).]Well how about showing the limit of [tex]a_n-a_{n+1}=0[/tex] as n->oo

What if a_n is bound both above and bellow?

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pivoxa15

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John Creighto said:What if a_n is bound both above and bellow?

In which case it would be in a compact space. All cauchy sequences in a compact space converge.

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morphism

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Just because (a_n - a_{n+1}) -> 0 doesn't mean a_n is Cauchy. Again, a counterexample to this claim is a_n = log(n). We have log(n) - log(n+1) = log(n/(n+1)) -> log(1) = 0, but if this were Cauchy, then it would be convergent (since the reals are complete), and it clearly isn't.pivoxa15 said:In which case it would be in a compact space. All cauchy sequences in a compact space converge.

It turns out that being bounded isn't good enough either, although finding a counterexample was trickier. At any rate: try a_n = exp(i(1 + 1/2 + ... + 1/n)). This doesn't converge - it goes around the unit circle in the complex plane. On the other hand,

[tex]a_n - a_{n+1} = \exp\left(i \left(1 + \frac{1}{2} + ... + \frac{1}{n}\right)\right)\left(1 - \exp\left(\frac{i}{n+1}\right)\right) \to 0.[/itex]

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John Creighto

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morphism said:Just because (a_n - a_{n+1}) -> 0 doesn't mean a_n is Cauchy. Again, a counterexample to this claim is a_n = log(n). We have log(n) - log(n+1) = log(n/(n+1)) -> log(1) = 0, but if this were Cauchy, then it would be convergent (since the reals are complete), and it clearly isn't.

It turns out that being bounded isn't good enough either, although finding a counterexample was trickier. At any rate: try a_n = exp(i(1 + 1/2 + ... + 1/n)). This doesn't converge - it goes around the unit circle in the complex plane. On the other hand,

[tex]a_n - a_{n+1} = \exp\left(i \left(1 + \frac{1}{2} + ... + \frac{1}{n}\right)\right)\left(1 - \exp\left(\frac{i}{n+1}\right)\right) \to 0.[/itex]

Okay, how about [tex]a_n-a_{n-1} \le k \left( \frac{1}{n}-\frac{1}{n-1} \right) [/tex] for all n greater then M

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- #13

John Creighto

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morphism said:It turns out that being bounded isn't good enough either, although finding a counterexample was trickier. At any rate: try a_n = exp(i(1 + 1/2 + ... + 1/n)). This doesn't converge - it goes around the unit circle in the complex plane. On the other hand,

[tex]a_n - a_{n+1} = \exp\left(i \left(1 + \frac{1}{2} + ... + \frac{1}{n}\right)\right)\left(1 - \exp\left(\frac{i}{n+1}\right)\right) \to 0.[/itex]

That's a pretty interesting example. Can you find a similar example on the reals?

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morphism

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What are k and M?John Creighto said:Okay, how about [tex]a_n-a_{n-1} \le k \left( \frac{1}{n}-\frac{1}{n-1} \right) [/tex] for all n greater then M

Take the real part of it and see what happens.That's a pretty interesting example. Can you find a similar example on the reals?

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pivoxa15

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To be more specific, I had a sequence a_m/b_m where a_m and b_m in the limit is 0.

However they are not functions so can't use that rule which I can't spell, l'hospil's rule?

However they are not functions so can't use that rule which I can't spell, l'hospil's rule?

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John Creighto

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morphism said:What are k and M?

Take the real part of it and see what happens.

k an M are arbitrary chosen to make the inequality work for a given sequence.

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HallsofIvy

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If all you know is that [itex]a_n[/itex] and [itex]b_n[/itex] go to 0, you can't say anything about whether [itex]a_n/b_n[/itex] converges or diverges. For example, if [itex]a_n= 1/n^2[/itex] and [itex]b_n= 1/n[/itex], it is obvious that both [itex]a_n[/itex] and [itex]b_n[/itex] converge to 0. And [itex]a_n/b_n= (1/n^2)(n/1)= 1/n[/itex] converges to 0.

But if [itex]b_n= 1/n^2[/itex] and [itex]a_n= 1/n[/itex], it is still obvious that both [itex]a_n[/itex] and [itex]b_n[/itex] converge to 0 but now [itex]a_n/b_n= (1/n)(n^2/1)= n[/itex] does not converge.

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ssd

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Looking upon the 1st post: if left hand and right hand limits are equal, then the limit exists.

The limit of a function is the value that a function approaches as the input (x) approaches a certain value (c). In other words, it is the value that the function is "approaching" but may never actually reach for a given input.

A limit is typically calculated using algebraic manipulation and substitution. For example, if the limit of a function as x approaches 3 is required, the function can be simplified and then the value of x can be substituted as 3 to find the limit.

A one-sided limit is when the function is approaching a specific value from only one direction, either from the left or the right. A two-sided limit is when the function is approaching a specific value from both the left and the right.

To prove the existence of a limit, you must show that the function approaches a specific value as the input approaches a certain value. This can be done through various methods, such as using the definition of a limit, algebraic manipulation, or graphical analysis.

Proving the existence of a limit is important because it allows us to make predictions about the behavior of a function and the values it will approach as the input changes. It also helps us to understand and analyze the behavior of more complex functions.

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