# The definition for limit in analysis

1. Jun 18, 2008

### Shing

Hi guys
I am quite confused for the definition
since it is some "simply" written and with lots of important information
e is the approximation

given e>0
| a(n)-L | < e
for n >> 1 ( for n large)

it says something " the approximation can be made as close as desired, provided we go far enough out in the sequence - the smaller e is, the farther out we must go, in general"

basically, I don't understand
1,) what mean by "the farther out we must go"

2.) How can I read such huge information from the mere there lines? Did I miss something?

3.) is " given" mean "when consider..."?

4.) and the whole information !!

2. Jun 18, 2008

### ModernLogic

1) "the smaller e is, farther out we must go" means the closer we want the sequence to get to its limit, the bigger we choose n to be. As an example, consider the sequence a(n)=1/n. The limit as n goes to infinity is zero. One way to prove this is to show that by selecting a large enough value of n (or "farther out we must go," as you put it), I can get as close to zero as required. So if I want a(n) to be within 0.01 of 0, namely the interval (-0.01,0.01), I can simply pick n to be greater than 100.

2) This is a vague and somewhat opinionated question. Math statements are often very succinct, containing lots of information in very few words. I think its rather eloquent.

3) Yes... I think. "Given e=0.01" could mean "consider when e=0.01" in the context of limits. Though, I personally would not put it that way.

4) The best way to understand the definition of limit is to think of it as an argument between yourself and a good friend, pertaining to the limit of a function. Going back to our example of a(n)=1/n, suppose your friend does not believe a(n)--> 0 as n--> infinity. You say: "well, I bet the sequence can get as close to 0 as possible." He says: "Can it get within 0.01 of its limit, 0?" You say: "Sure, pick n>100." He says: "Alright then, can it get within 0.0001 of its limit, 0?" Again, you respond with: "Yes, for n>10000." And so on hence proving that the limit is in fact 0.

3. Jun 18, 2008

### tiny-tim

Hi Shing!

(have an epsilon: ε )

The statement "the limit of a(n) as n tends to ∞ is L"

is defined as meaning:

Consider any ε > 0, however small: then there is an N(ε) (which depends on ε) such that | a(n)-L | < ε for all n > N(ε).

"the farther out we must go" means that, as ε gets smaller, N(ε) must get larger.

4. Jun 19, 2008

### Shing

let me express my understanding..

given $$\epsilon>0$$
that implies the limit is NOT equal to the sequence no matter how larger n is.

$$|a_n-L|<\epsilon ,$$ for $$n >> 1$$
that implies the difference between $a_n$ and L is within a very small value$\epsilon$ when n is large enough.

What is the problem of my concept?
Can I get better understanding of the definition?

5. Jun 19, 2008

### LukeD

Not necessarily.

Consider the sequence <1,1,1,1,1,...>
The limit is 1, and every term of the sequence is also 1.

Also consider the sequence <1, 1/2, 1/3, 1/4, 1/5, ...>
The limit is 0, but every term of the sequence is greater than 0 (so 0 is never actually encountered)

<1, 0, 1, 2, 3, 10, 253, 0, 0, 0, 0, 0, ...>
After a finite number of terms every subsequent term is 0, so the limit is 0

<0, 1, 0, -1, 0, 1/2, 0, -1/2, 0, 1/3, 0, -1/3, 0, 1/4, 0, -1/4, ...>
As you go "further out" each of the terms get arbitrarily close to 0. Many of the terms are in fact 0, many of them are not. However, because they get arbitrarily close to 0, the limit is 0.

But consider the sequence
<0, 1, 0, 0, 2, 0, 0, 0, 3, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 5, ...> (the the pattern being n zeros followed by n, followed by n+1 zeros followed by n+1 and so on)
Although the frequency of 0 is increasing, the terms are not in general getting arbitrarily close to any value, so there is no limit.

6. Jun 20, 2008

### HallsofIvy

Staff Emeritus
That's simply not a very good definition. Instead of "for n >> 1 ( for n large)" most texts have something like "there exist N such that if n> N, then..." If it is possible to find such an n no matter what $\epsilon$ is, then the limit is L. As for "the farther we must go, notice it said "in general". That is not always true but the situations in which it is not are generally simple.

Not necessarily. For example the sequence an= 1, a "constant sequence" has limit 1 (of course) and is always equal to 1. In that case |an-L|= |1-1|= 0 < $\epsilon$ for all $\epsilon$ and all n.

7. Jun 26, 2008

### Shing

So
what does
$$given ε>0$$
tell me?

It tells me
$$|a_n-L|<\epsilon , for n >> N$$
tells me:
Can I deepen my understanding of limit?

And for another definition of limit I found is much more readable.

Am I correct? ( because I changed it a bit from something about matrix space.)

8. Jun 26, 2008

### HallsofIvy

Staff Emeritus
I see no difference between the two definitions. In the real numbers, the standard "distance" measure is $d(p_n, P)= |p_n- P|$ so they say exactly the same thing.

9. Jun 26, 2008

### tiny-tim

best definition

Hi Shing!

That last definition is the best.

I would change it slightly, to:

For any ε > 0, there is an integer N such that |pn - P| < ε for all n > N.​