Limit of a sequence (explanation)

[Korisnik]

I kind of know what limits are, or at least believe I do: I think that a limit of a sequence is just an approximation/intuitive way to finding a number (if it exists) to which a sequence tends. For example, 1, 2, 3, 4... tends to +∞, while 1/10, 1/100, 1/1000... tends, "obviously/intuitively", to 0.

Now formal definition of a limit of a sequence in my book is:

A sequence of real numbers a_n converges (tends) to a real number L if for all \varepsilon>0 exists n_0\in\mathbb{N} such that for all n\geq n_0 it's true |a_n-L|<\varepsilon. Then L is limit of a sequence a_n,
L=\lim_{n\to \infty} a_n.
Or symbolically: \lim_{n\to\infty}a_n=L\Longleftrightarrow(\forall\varepsilon>0)(\exists n_0\in\mathbb{N})(\forall n\geq n_0)(|a_n-L|<\varepsilon).

I understand the first part of the definition: for every \varepsilon>0 there is n_0\in\mathbb{N} such that for every n\geq n_0 it's true...

This means that we first pick a epsilon bigger than 0, then there must be (if the limit exists) some n_0 from which (to infinity) all members will be of a greater ordinal number -- and they (\varepsilon we picked and n_0 that corresponds) comply with the condition |a_n-L|<\varepsilon.

Now this condition, the inequality with the absolute value represents a interval:
-\varepsilon<a_n-L<\varepsilon\Longleftrightarrow a_n\in(L-\varepsilon, L+\varepsilon).

This, I think, means that a_n (as n\to\infty) is near the limit L value, but in the range of left and right boundaries of \varepsilon.

Represented (poorly) by a number line:

-∞ · · · ------(-ε) a_n-L (ε)------ · · · +∞

From the equations I now think that as n approaches infinity, the a_n approaches limit (obviously), and that makes sense -- and it agrees with the equation (because if a_n reaches L, they will subtract to a 0, and since epsilon basically tends to a 0, that's the definition of a zero (-0.0000...1 < 0 < 0.0000...1)).

Finally, limit of a sequence is just an approximation of the "infinitieth" member of a sequence.

Tell me if I have understood anything wrongly. Thanks in advance.

 

[xAxis]

Yes, seems to me you nailed it. Someone might make problem about your last sentence about "infinitieth member", but don't pay much attention to them :).
There are few other corollaries involving "all but finitely many members of sequence being out of interval" which will help solidify your understanding.

 

[jbriggs444]

I kind of know what limits are, or at least believe I do: I think that a limit of a sequence is just an approximation/intuitive way to finding a number (if it exists) to which a sequence tends. For example, 1, 2, 3, 4... tends to +∞, while 1/10, 1/100, 1/1000... tends, "obviously/intuitively", to 0.
[...]
Finally, limit of a sequence is just an approximation of the "infinitieth" member of a sequence.

Intuitively, this is a decent way to think about limits. However the term "approximation" is not apt. A limit is not an approximation. It is an exact number.

The "infinitieth" member, if it were to exist and if the sequence were to be continuous at infinity would have to be exactly equal to the limit.

 

[mal4mac]

Yes, seems to me you nailed it. Someone might make problem about your last sentence about "infinitieth member", but don't pay much attention to them :).

.. unless it's your professor marking you down for it :)

 

[Stephen Tashi]

and since epsilon basically tends to a 0.

It's tempting to introduce some idea of motion into the intuitive idea of a limit since the formal definition uses the word "approaches". But in the formal definition, there isn't any motion or dynamic process. Epsilon isn't going anywhere. It isn't "tending" to anything. The freedom of choice guarnteed by the phrase "for each epsilon greater than 0" mean's you could pick a sequence of epsilons that "tended" to zero, but doing this is not specified in the definition.

Understanding the formal definition of limit involves comprehending how it avoids using concepts of dynamic motion ("tending", "approaching") by clever use of the logical quantifiers "for each" and "there exists".

 

[economicsnerd]

OP, everything you put looks basically right to me.

A bit of terminology that might be helpful for phrasing this stuff with less notation: the word "eventually". Given a property "blahblahblah", say a sequence $(x_n)_{n=1}^\infty$ is eventually blahblahblah if there exists a number $n_0\in\mathbb N$ such that for every $n\in\mathbb N$ with $n\geq n_0$, the number $x_n$ is blahblahblah.

For example, the sequence with $x_n = n - 72304551$ is eventually positive, as we can see by setting $n_0 = 72304552$. The sequence of integers with $x_n= \begin{cases} 12 & \text{ if n is a power of } 10, \\ 7 & \text{ otherwise} \end{cases}$ isn't eventually odd, because there are arbitrarily late members of the sequence that fail to be odd.

Now limits become even easier to describe:
- $(x_n)_n$ converges to $x\in\mathbb R$ if for every $\epsilon>0,$ the sequence is eventually in $(x-\epsilon, x+\epsilon)$.
- $(x_n)_n$ converges to $\infty$ if for every $K\in\mathbb R$ the sequence is eventually bigger than $K$.

 

[Korisnik]

Intuitively, this is a decent way to think about limits. However the term "approximation" is not apt. A limit is not an approximation. It is an exact number.

Yes, I expressed myself badly (english isn't my mother tongue), what I meant was -- a focus of a better and better approximation?

economicsnerd, thank You for the tips.

It's tempting to introduce some idea of motion into the intuitive idea of a limit since the formal definition uses the word "approaches". But in the formal definition, there isn't any motion or dynamic process. Epsilon isn't going anywhere. It isn't "tending" to anything. The freedom of choice guarnteed by the phrase "for each epsilon greater than 0" mean's you could pick a sequence of epsilons that "tended" to zero, but doing this is not specified in the definition.

Understanding the formal definition of limit involves comprehending how it avoids using concepts of dynamic motion ("tending", "approaching") by clever use of the logical quantifiers "for each" and "there exists".

Thank You very much for this post, you have a really good point!

When I wrote the sentence you quoted, I myself didn't agree with it. Somehow I realized I could've written that at the beginning and then everything would be easier to explain; but I didn't, because I wasn't sure in it, it wasn't "my original idea"... Though it made more sense, it was just so simple... like an oversimplification of something and abstraction to a broader concept so it basically lost it's meaning (my meaning, my thinking of the problem). Your post made me at the beginning again. Thank You for that.

I think I'll have to sleep on the idea you mentioned and to think through it. But I'll write down my first thoughts of the definition for now: first it defines L as a number, so one number, not an interval. This will be very important... Now they could've just made this condition |a_n-L|=0 but then that could be any number, what exactly are we looking for? We are looking for one number. By clever use of: for all \varepsilon exist all n that satisfy the condition -- mentally plugging in all possible numbers for \varepsilon and a_n -- it's obvious that all choices of \varepsilon result in a range of a_n members (there is always a n_0, such that n\geq n_0), except for the smallest \varepsilon possible (a number "next to" 0), for which exists only one member of a_n: the last one. This last member is just n_0, one number, because if there were a next number, it would be smaller than 0 and would add to L, that would be bigger than 0 which is impossible by the condition. Now that we've eliminated all other members of the sequence, we are left with a number that is in this range -0.0...1 < a_n-L < 0.0...1. And what number is between these infinitely small numbers? It's 0. This is where the definition ends. Because now we can say that a_{n=last}=L, because if a-b=0 then a=b.