# Limit of A Sequence

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• Nov 15th 2009, 11:11 PM
RockHard
Limit of A Sequence
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
• Nov 16th 2009, 03:46 AM
rubix
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.
• Nov 16th 2009, 04:24 AM
tonio
Quote:

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
• Nov 16th 2009, 10:16 AM
RockHard
How would find just some "random" N I understand your first statement of choosing some n close to the limit
• Nov 16th 2009, 10:37 AM
Defunkt
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.
• Nov 16th 2009, 12:59 PM
RockHard
Hm, thank sort of make sense. Perhaps a basic problem could help me visualize this
• Nov 16th 2009, 01:31 PM
redsoxfan325
Quote:

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.
• Nov 16th 2009, 01:51 PM
RockHard
Quote:

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
• Nov 16th 2009, 02:03 PM
redsoxfan325
Quote:

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.
• Nov 16th 2009, 03:38 PM
RockHard
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 (Happy)
• Nov 16th 2009, 03:45 PM
Drexel28
Quote:

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

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

matter?

What do you mean here? If you mean does it matter whether it is $\left|a_n-L\right|<\varepsilon$ or $\left|L-a_n\right\|<\varepsilon$ the answer is no. This is because $\left|a_n-L\right|=\left|-\left(L-a_n\right)\right|=|-1|\left|L-a_n\right|=\left|L-a_n\right|$. In other words, absolute values dont care about signs. If you meant does it matter whether it is $\left|a_n-L\right|<\varepsilon$ or $\left|a_n-L\right|>\varepsilon$ the answer is...absolutely. One is saying you can make $a_n$ arbitrarily close...the other is saying, well that you can make their difference arbitrarily large.
• Nov 16th 2009, 03:47 PM
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, AHHHHHHHH nevermind I understand what he was proving
• Nov 16th 2009, 03:53 PM
Drexel28
Quote:

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?
• Nov 16th 2009, 03:55 PM
RockHard
Yes very much so thanks, going for another example to do
• Nov 16th 2009, 05:01 PM
RockHard
Can anyone provide me with another example before I deem this thread solved
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