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Math Help - Limit of A Sequence

  1. #1
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    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
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  2. #2
    Junior Member rubix's Avatar
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    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.
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  3. #3
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    Quote Originally Posted by RockHard View Post
    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
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    How would find just some "random" N I understand your first statement of choosing some n close to the limit
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  5. #5
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    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.
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  6. #6
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    Hm, thank sort of make sense. Perhaps a basic problem could help me visualize this
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  7. #7
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by RockHard View Post
    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.
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  8. #8
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    Quote Originally Posted by redsoxfan325 View Post
    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
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  9. #9
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by RockHard View Post
    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.
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  10. #10
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    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
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  11. #11
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by RockHard View Post
    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.
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  12. #12
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    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
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  13. #13
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by RockHard View Post
    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?
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  14. #14
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    Yes very much so thanks, going for another example to do
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  15. #15
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    Can anyone provide me with another example before I deem this thread solved
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