# I Why this limit exists?

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1. Oct 23, 2016

### Buffu

$$x_n = (-1)^n {2n\over n+1} \sin n$$

it is given that ,

$$|x_n| = |(-1)^n| {2n\over n+1} |\sin n| < {2n\over n+1} < 2$$
thus bounded, but what i did not get is how did we find $\lim_{n \to \infty} |(-1)^n| {2n\over n+1} |\sin n| = 2$.
I checked with wolfram alpha and it says $\lim_{n \to \infty} |(-1)^n| {2n\over n+1} |\sin n| = 2$.
https://www.wolframalpha.com/input/?i=lim+n+to+infinity+(-1)^n+*+sin+n+*+(2n/(n+1))

My try

$$\lim_{n \to \infty} |(-1)^n| {2n\over n+1} |\sin n|$$
$$\lim_{n \to \infty} {|(-1)^n| \over n} {2n^3\over n+1} {|\sin n| \over n}$$
$$\lim_{n \to \infty} {|(-1)^n| \over n}{2\over {1\over n^2}+{1\over n^3}} {|\sin n |\over n} ={2 \over 0}$$

2. Oct 23, 2016

### andrewkirk

The limit does not exist, and it's not clear to me that Wolfram is saying it does. The expression on the results page is vague but there is certainly no clear statement that the limit exists.

In fact, for any $N\in\mathbb N$ and $\epsilon>0$ we can find $m>N$ such that $\bigg||x_m|-0\bigg|<\epsilon$.

3. Oct 23, 2016

### pwsnafu

4. Oct 23, 2016

### Buffu

So a sequence can diverge and still be bounded ?

5. Oct 23, 2016

### Buffu

Thanks.
But can you help me understand how does the given sequence is bounded ?
It clearly does diverge, so should be a unbounded sequence, right ?

6. Oct 23, 2016

### Staff: Mentor

Yes. A very simple sequence that is bounded and that diverges is $s_n = (-1)^n = \{1, -1, 1, -1, +- \dots \}$.

7. Oct 23, 2016

### Staff: Mentor

No. $((-1)^n)_{n \in \mathbb{N}}$ clearly diverges because it does not converge. On the other hand it is clearly bounded by $-1 \leq (-1)^n \leq 1$

8. Oct 23, 2016

### Buffu

Thanks.
That cleared many of my doubts.

9. Oct 24, 2016

### Svein

There are two concepts here which are different but similar.
• w is a cluster point for a sequence {xn} if for any ε>0 and any N you can find an m>N such that |xm-w|<ε.
• z is a limit point for a sequence {xn} if for any ε>0 you can find an N such that for all m>N you have |xm-w|<ε.
Thus if a sequence has only one cluster point, that cluster point is a limit point.

For the sequence in the OP, every point in [-2, 2] is a cluster point. BUT there is no limit point.

10. Oct 24, 2016

### Staff: Mentor

By this definition, the sequences $s_n = \{(-1)^n\}$ and $\sigma_n = \{2 + (-1)^n\frac n {n + 1}\}$ have no cluster points. No matter how large N is, there will be some members of each sequence that are farther from w than ε. In other words, not all elements in the tail of the sequence are close to some limiting value.
Edit: I misinterpreted what Svein was saying in the definition above. He's not saying that |xm - w| < ε for every m > N, just some m > N.

However, each of these sequences has a subsequence that converges.
For the sequence $s_n$ above, the subsequence $\{s_0, s_2, s_4, \dots, s_{2n}, \dots \} = \{1, 1, 1, \dots, 1, \dots \}$ is convergent, so 1 is a cluster point. Likewise, the subsequence of terms with odd indexes is $\{-1, -1, -1, \dots, -1, \dots \}$ is convergent, so -1 is also a cluster point. For the sequence itself, there is no limit point, hence no limit.

For the sequence $\sigma_n$ above, there are two convergent subsequences - the subsequence made up of terms with even indexes $\{2, 2 + 2/3, 2 + 4/5, \dots, 2 + n/(n + 1), \dots \}$, which converges to 3. There is also the subsequence of terms with odd indexes, which converges to 1. This sequence has two cluster points, 3 and 1. For the sequence itself, there is no limit point, hence no limit.

Last edited: Oct 24, 2016
11. Oct 24, 2016

### Krylov

I think you may have misread something here? You only need to find one $m > N$.

12. Oct 24, 2016

### Staff: Mentor

I'm interpreting what Svein said to mean that for a given N, and m > N, |xm - w| < ε.

13. Oct 24, 2016

### Krylov

Fine. That is not what he wrote, though.

14. Oct 24, 2016

### Staff: Mentor

I edited my reply to Svein.

15. Oct 24, 2016

### mathman

$\lim_{n \to \infty}|(-1)^n\frac{2n}{n+1}|=2$. However |sin(n)| oscillates between 0 and 1, so there is no limit for the expression.

16. Oct 25, 2016

### Svein

That is why English is badly suited to mathematics. I shall rephrase using math symbols:
• w is a cluster point for a sequence {xn} if (∀ε>0)(∀N>0)(∃m>N)(|xm-w|<ε).
• z is a limit point for a sequence {xn} if (∀ε>0)(∃N>0)(∀m>N)(|xm-z|<ε).

17. Oct 25, 2016

### zinq

"w is a cluster point for a sequence {xn} if for any ε>0 and any N you can find an m>N such that |xm-w|<ε."

Yes!

"z is a limit point for a sequence {xn} if for any ε>0 you can find an N such that for all m>N you have |xm-w|<ε."

No!

In fact the phrase "limit point" is always used to mean exactly the same thing as your definition of cluster point above. The phrase "limit point" does not mean "limit". Although there are certainly some cases where limit points are also limits.

The correct terminology for "limit" in English is the word "limit" (not "limit point"). (And beware: Occasionally the term "cluster point" is used to mean something other than "limit point". I have seen "cluster point" used for a point (of a set) every neighborhood of which contains *uncountably* many points of the set. But this is not standard.)

NOTE: The term "limit" is appropriate for a sequence, but the term "limit point" is appropriate to apply to a set.

My apologies if this may be confusing, but that's how the words are used.

18. Oct 25, 2016

### Svein

Well, I can tentatively agree with you on the phrase "limit point" - I should have written "limit". Otherwise, my definitions are identical to the ones used in Royden: Real Analysis. The usage "limit point" may be more appropriate in a topological context.

19. Oct 26, 2016

### andrewkirk

Yes I think the term 'limit point' is more often used in topology. I also found it confusing when I first encountered it, as it is quite different from the usual meaning of 'limit' in analysis.

I prefer the term 'accumulation point', as it doesn't risk generating the same confusion that the word 'limit' brings. Or we could say a set $S\subset\mathbb R$ is 'dense' at point $P\in\mathbb R$.

A sense in which a limit point $P$ of a subset $S$ of a metric space is a genuine limit is that there exists a sequence of elements of $S-\{P\}$ whose limit is $\{P\}$. We may however need to use the Axiom of Choice to assert the existence of such a sequence, unless we have some pre-existing structure that enables us to identify the sequence without using Choice.

Another interesting (to me) observation is that a point being a limit of a sequence does not necessarily make it a limit point of the set comprising the elements of the sequence. For instance, the point 1 is the limit, but not a limit point of the associated set, of the constant sequence 1,1,1,1...... In fact that set {1} has no limit points in $\mathbb R$.

20. Oct 27, 2016

### Buffu

I am lost, can you please explain me the difference between point and limit in easy words? please.