Limit of A Sequence

**RockHard** · November 15th 2009, 10:11 PM

I am having trouble grasping the concept of a limit in terms of sequences as I am trying to learn sequences/series. I feel this is an important concept to be grasped but the definition of the limit is so abstract I can understand for any epsilion greater than 0, meaning any number someone gives you > 0 exists some N, whenever
|an - L| < epsilion and n > N
I can kinda see where you get the nth term but where does some random N come from where is it determined in this defintion and the last bit I mentioned too is a bit abstract

**rubix** · November 16th 2009, 02:46 AM

yes, limit is a VERY important topic.

the uppercase N is some natural number and n is some large number bigger than N that satisfies $|a_n-L| < \epsilon$

---

the formal defination goes like this:

a sequence $(a_n)$ has a limit L if, for every $\epsilon > 0$ there exists a Natural number N such that $|a_n - L| < \epsilon$ whenever $n > N$

i.e. (a_n) -> $L => \forall \epsilon > 0, \exists N \in \mathbb{N}:$

$|a_n - L| < \epsilon whenever n > N$

essentially what you are doing is making the sequence $(a_n)$ as close to its limit L by taking n large enough. This lowercase n has to be bigger than some uppercase N which is a natural number.

Hope that helps!

if you have any qn post back.

**tonio** · November 16th 2009, 03:24 AM

Originally Posted by RockHard

I am having trouble grasping the concept of a limit in terms of sequences as I am trying to learn sequences/series. I feel this is an important concept to be grasped but the definition of the limit is so abstract I can understand for any epsilion greater than 0, meaning any number someone gives you > 0 exists some N, whenever
|an - L| < epsilion and n > N
I can kinda see where you get the nth term but where does some random N come from where is it determined in this defintion and the last bit I mentioned too is a bit abstract

You could read the first tutorial at the head of the Calculus page by Krizalid about evaluating limits according to the definition. I think it may help you.

Tonio

**RockHard** · November 16th 2009, 09:16 AM

How would find just some "random" N I understand your first statement of choosing some n close to the limit

**Defunkt** · November 16th 2009, 09:37 AM

The N you choose isn't random, it depends on $\epsilon$ -- think of it like this: the smaller $\epsilon$ is, the larger your N will be, since you will need to go "further" in the sequence to get closer and closer to the limit... That is, unless $a_n$ is constant for all n.

**RockHard** · November 16th 2009, 11:59 AM

Hm, thank sort of make sense. Perhaps a basic problem could help me visualize this

**redsoxfan325** · November 16th 2009, 12:31 PM

Originally Posted by RockHard

Hm, thank sort of make sense. Perhaps a basic problem could help me visualize this

Prove, using the definition of limit, that the sequence $a_n=5-\frac{3}{n+10}$ converges to $5$ as $n\to\infty$ .

This is a nice simple problem that will test your basic understanding of the concept.

**RockHard** · November 16th 2009, 12:51 PM

Originally Posted by redsoxfan325

Prove, using the definition of limit, that the sequence $a_n=5-\frac{3}{n+10}$ converges to $5$ as $n\to\infty$ .

This is a nice simple problem that will test your basic understanding of the concept.

Well you can use the Theorem that says you can treat the function a(n) as a function of x to help determine the limit so

The limit as $x\to\infty$
$5-\frac{3}{x+10}$

Similarly by properties of limits we can do

The limit as for $x\to\infty$
$5$ - $\frac{3}{n+10}$
separately

The limit of 5 is obviously just 5 because it is constant and then the limit of the second portion is as such

$\frac{3}{n+10}$

$\frac{\frac{3}{n}}{\frac{n}{n}+\frac{10}{n}}$

which then each indiviual limit is taking your whole limit is equal to

$\frac{0}{1}$ which is obviously 0

then you can combined the to which is again...obviously the limit of 5 which is 5

Hope I did this right or I feel incredible dumb, I don't if i used the definition properly by the terms discussed above

**redsoxfan325** · November 16th 2009, 01:03 PM

Originally Posted by RockHard

Well you can use the Theorem that says you can treat the function a(n) as a function of x to help determine the limit so

The limit as $x\to\infty$
$5-\frac{3}{x+10}$

Similarly by properties of limits we can do

The limit as for $x\to\infty$
$5$ - $\frac{3}{n+10}$
separately

The limit of 5 is obviously just 5 because it is constant and then the limit of the second portion is as such

$\frac{3}{n+10}$

$\frac{\frac{3}{n}}{\frac{n}{n}+\frac{10}{n}}$

which then each indiviual limit is taking your whole limit is equal to

$\frac{0}{1}$ which is obviously 0

then you can combined the to which is again...obviously the limit of 5 which is 5

Hope I did this right or I feel incredible dumb, I don't if i used the definition properly by the terms discussed above

Your reasoning is all fine, but to use the definition of limit we need to use the $\epsilon-N$ argument.

We want to show that $\forall~\epsilon>0$ , $\exists N$ such that $n>N$ implies

$|L-a_n|=\left|5-\left(5-\frac{3}{n+10}\right)\right|<\epsilon$

$\left|5-\left(5-\frac{3}{n+10}\right)\right|=\frac{3}{n+10}$

To solve $\frac{3}{n+10}<\epsilon$ , we get $\frac{n+10}{3}>\frac{1}{\epsilon}\implies n+10>\frac{3}{\epsilon}\implies n>\frac{3}{\epsilon}-10$

Now that the algebra is done, the actual proof is quite simple.

Proof: Given an $\epsilon>0$ , let $N=\frac{3}{\epsilon}-10$ . So for $n>N$ ,

$\left|5-\left(5-\frac{3}{n+10}\right)\right|=\frac{3}{n+10}<\frac{ 3}{(\frac{3}{\epsilon}-10)+10}=\frac{3}{3/\epsilon}=\epsilon$

and we're done.

Note: Most professors won't make you show the work to actually find what $N$ should be. My experience has been that you can just assert what $N$ should be, and as long as it works then it's fine.

**RockHard** · November 16th 2009, 02:38 PM

Hmm it is slowly starting to make some sense, does the order of this statement

$|{a_n - L}| <\epsilon$

matter? I still making to sense the last part you did using the proof the statements above are slowly starting to formulate, you just find the limit set to

$|{a_n - L}| <\epsilon$

and solve for n and I think for the last bit you plugged in what you solved for n > some epsilon in the last bit

Thanks for your time bro, everyone gets reps and thanks after this

**Drexel28** · November 16th 2009, 02:45 PM

Originally Posted by RockHard

Hmm it is slowly starting to make some sense, does the order of this statement

$|{a_n - L}| <\epsilon$

matter?

**RockHard** · November 16th 2009, 02:47 PM

Thanks man!
Also for his last part where he I assume is solving for epsilon what does that statement exactly mean, I understand the N = statement, AHHHHHHHH nevermind I understand what he was proving

**Drexel28** · November 16th 2009, 02:53 PM

Originally Posted by RockHard

Thanks man!
Also for his last part where he I assume is solving for epsilon what does that statement exactly mean, I understand the N = statement

If I am interpreting your question correctly, you mean "what does his "find the N value" mean?". The concept of $\lim_{n\to\infty}a_n=L$ is that just like was said above I can make $a_n$ and $L$ as arbitrarily close (since it's arbitrary let's call that distance epsilon) by making $n$ sufficiently large. In other words if you give me ANY $\varepsilon>0$ (it really does not matter what it is) I need to find some $N\in\mathbb{N}$ such that $n\ge N\implies \left|a_n-L\right|<\varepsilon$ . So if a teacher asked you to prove that $\lim_{n\to\infty}\frac{n}{n+1}=1$ what it would mean is that if I give you ANY $\varepsilon>0$ (think of it like a variable...but one that is fixed under specific cases) you can tell me the EXACT $N$ such that $n\ge N\implies \left|\frac{n}{n+1}-1\right|<\varepsilon$ . Does that help?

**RockHard** · November 16th 2009, 02:55 PM

Yes very much so thanks, going for another example to do

**RockHard** · November 16th 2009, 04:01 PM

Can anyone provide me with another example before I deem this thread solved

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