text: # Limit of A Sequence
  
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  • Nov 15th 2009, 10:11 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, 02:46 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 $\displaystyle |a_n-L| < \epsilon$
  
  ---
  
  the formal defination goes like this:
  
  a sequence $\displaystyle (a_n)$ has a limit L if, for every $\displaystyle \epsilon > 0$ there exists a Natural number N such that $\displaystyle |a_n - L| < \epsilon$ whenever $\displaystyle n > N$
  
  i.e. (a_n) -> $\displaystyle L => \forall \epsilon > 0, \exists N \in \mathbb{N}:$
  
  $\displaystyle |a_n - L| < \epsilon whenever n > N$
  
  essentially what you are doing is making the sequence $\displaystyle (a_n)$ 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, 03:24 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, 09:16 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, 09:37 AM
  Defunkt
  The N you choose isn't random, it depends on $\displaystyle \epsilon$ -- think of it like this: the smaller $\displaystyle \epsilon$ 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 $\displaystyle a_n$ is constant for all n.
  • Nov 16th 2009, 11:59 AM
  RockHard
  Hm, thank sort of make sense. Perhaps a basic problem could help me visualize this
  • Nov 16th 2009, 12:31 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 $\displaystyle a_n=5-\frac{3}{n+10}$ converges to $\displaystyle 5$ as $\displaystyle n\to\infty$.
  
  This is a nice simple problem that will test your basic understanding of the concept.
  • Nov 16th 2009, 12:51 PM
  RockHard
  Quote:
  
  Originally Posted by redsoxfan325
  Prove, using the definition of limit, that the sequence $\displaystyle a_n=5-\frac{3}{n+10}$ converges to $\displaystyle 5$ as $\displaystyle n\to\infty$.
  
  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 $\displaystyle x\to\infty$
  $\displaystyle 5-\frac{3}{x+10}$
  
  Similarly by properties of limits we can do
  
  The limit as for $\displaystyle x\to\infty$
  $\displaystyle 5$ - $\displaystyle \frac{3}{n+10}$
  separately
  
  The limit of 5 is obviously just 5 because it is constant and then the limit of the second portion is as such
  
  $\displaystyle \frac{3}{n+10}$
  
  $\displaystyle \frac{\frac{3}{n}}{\frac{n}{n}+\frac{10}{n}}$
  
  which then each indiviual limit is taking your whole limit is equal to
  
  $\displaystyle \frac{0}{1}$ 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, 01:03 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 $\displaystyle x\to\infty$
  $\displaystyle 5-\frac{3}{x+10}$
  
  Similarly by properties of limits we can do
  
  The limit as for $\displaystyle x\to\infty$
  $\displaystyle 5$ - $\displaystyle \frac{3}{n+10}$
  separately
  
  The limit of 5 is obviously just 5 because it is constant and then the limit of the second portion is as such
  
  $\displaystyle \frac{3}{n+10}$
  
  $\displaystyle \frac{\frac{3}{n}}{\frac{n}{n}+\frac{10}{n}}$
  
  which then each indiviual limit is taking your whole limit is equal to
  
  $\displaystyle \frac{0}{1}$ 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 $\displaystyle \epsilon-N$ argument.
  
  We want to show that $\displaystyle \forall~\epsilon>0$, $\displaystyle \exists N$ such that $\displaystyle n>N$ implies
  
  $\displaystyle |L-a_n|=\left|5-\left(5-\frac{3}{n+10}\right)\right|<\epsilon$
  
  $\displaystyle \left|5-\left(5-\frac{3}{n+10}\right)\right|=\frac{3}{n+10}$
  
  To solve $\displaystyle \frac{3}{n+10}<\epsilon$, we get $\displaystyle \frac{n+10}{3}>\frac{1}{\epsilon}\implies n+10>\frac{3}{\epsilon}\implies n>\frac{3}{\epsilon}-10$
  
  Now that the algebra is done, the actual proof is quite simple.
  
  Proof: Given an $\displaystyle \epsilon>0$, let $\displaystyle N=\frac{3}{\epsilon}-10$. So for $\displaystyle n>N$,
  
  $\displaystyle \left|5-\left(5-\frac{3}{n+10}\right)\right|=\frac{3}{n+10}<\frac{ 3}{(\frac{3}{\epsilon}-10)+10}=\frac{3}{3/\epsilon}=\epsilon$
  
  and we're done.
  
  Note: Most professors won't make you show the work to actually find what $\displaystyle N$ should be. My experience has been that you can just assert what $\displaystyle N$ should be, and as long as it works then it's fine.
  • Nov 16th 2009, 02:38 PM
  RockHard
  Hmm it is slowly starting to make some sense, does the order of this statement
  
  $\displaystyle |{a_n - L}| <\epsilon$
  
  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
  
  $\displaystyle |{a_n - L}| <\epsilon$
  
  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, 02:45 PM
  Drexel28
  Quote:
  
  Originally Posted by RockHard
  Hmm it is slowly starting to make some sense, does the order of this statement
  
  $\displaystyle |{a_n - L}| <\epsilon$
  
  matter?
  
  What do you mean here? If you mean does it matter whether it is $\displaystyle \left|a_n-L\right|<\varepsilon$ or $\displaystyle \left|L-a_n\right\|<\varepsilon$ the answer is no. This is because $\displaystyle \left|a_n-L\right|=\left|-\left(L-a_n\right)\right|=|-1|\left|L-a_n\right|=\left|L-a_n\right|$. In other words, absolute values dont care about signs. If you meant does it matter whether it is $\displaystyle \left|a_n-L\right|<\varepsilon$ or $\displaystyle \left|a_n-L\right|>\varepsilon$ the answer is...absolutely. One is saying you can make $\displaystyle a_n$ arbitrarily close...the other is saying, well that you can make their difference arbitrarily large.
  • Nov 16th 2009, 02:47 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, 02:53 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 $\displaystyle \lim_{n\to\infty}a_n=L$ is that just like was said above I can make $\displaystyle a_n$ and $\displaystyle L$ as arbitrarily close (since it's arbitrary let's call that distance epsilon) by making $\displaystyle n$ sufficiently large. In other words if you give me ANY $\displaystyle \varepsilon>0$ (it really does not matter what it is) I need to find some $\displaystyle N\in\mathbb{N}$ such that $\displaystyle n\ge N\implies \left|a_n-L\right|<\varepsilon$. So if a teacher asked you to prove that $\displaystyle \lim_{n\to\infty}\frac{n}{n+1}=1$ what it would mean is that if I give you ANY $\displaystyle \varepsilon>0$ (think of it like a variable...but one that is fixed under specific cases) you can tell me the EXACT $\displaystyle N$ such that $\displaystyle n\ge N\implies \left|\frac{n}{n+1}-1\right|<\varepsilon$. Does that help?
  • Nov 16th 2009, 02:55 PM
  RockHard
  Yes very much so thanks, going for another example to do
  • Nov 16th 2009, 04:01 PM
  RockHard
  Can anyone provide me with another example before I deem this thread solved
  Show 40 post(s) from this thread on one page
  Page 1 of 3 123 Last
explanation: The text extract is a discussion thread from a math forum where users are discussing the concept of limits in sequences. The thread starts with a user named RockHard expressing difficulty in understanding the formal definition of the limit of a sequence. Other users, such as rubix, tonio, and redsoxfan325, provide explanations and examples to help RockHard grasp the concept. The thread includes a detailed example of proving the limit of a sequence using the epsilon-N definition.
  
  The document contains a question and multiple answers to this question.
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distilabel_metadata.content: # Limit of A Sequence
  
  Show 40 post(s) from this thread on one page
  Page 1 of 3 123 Last
  • Nov 15th 2009, 10:11 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, 02:46 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 $\displaystyle |a_n-L| < \epsilon$
  
  ---
  
  the formal defination goes like this:
  
  a sequence $\displaystyle (a_n)$ has a limit L if, for every $\displaystyle \epsilon > 0$ there exists a Natural number N such that $\displaystyle |a_n - L| < \epsilon$ whenever $\displaystyle n > N$
  
  i.e. (a_n) -> $\displaystyle L => \forall \epsilon > 0, \exists N \in \mathbb{N}:$
  
  $\displaystyle |a_n - L| < \epsilon whenever n > N$
  
  essentially what you are doing is making the sequence $\displaystyle (a_n)$ 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, 03:24 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, 09:16 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, 09:37 AM
  Defunkt
  The N you choose isn't random, it depends on $\displaystyle \epsilon$ -- think of it like this: the smaller $\displaystyle \epsilon$ 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 $\displaystyle a_n$ is constant for all n.
  • Nov 16th 2009, 11:59 AM
  RockHard
  Hm, thank sort of make sense. Perhaps a basic problem could help me visualize this
  • Nov 16th 2009, 12:31 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 $\displaystyle a_n=5-\frac{3}{n+10}$ converges to $\displaystyle 5$ as $\displaystyle n\to\infty$.
  
  This is a nice simple problem that will test your basic understanding of the concept.
  • Nov 16th 2009, 12:51 PM
  RockHard
  Quote:
  
  Originally Posted by redsoxfan325
  Prove, using the definition of limit, that the sequence $\displaystyle a_n=5-\frac{3}{n+10}$ converges to $\displaystyle 5$ as $\displaystyle n\to\infty$.
  
  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 $\displaystyle x\to\infty$
  $\displaystyle 5-\frac{3}{x+10}$
  
  Similarly by properties of limits we can do
  
  The limit as for $\displaystyle x\to\infty$
  $\displaystyle 5$ - $\displaystyle \frac{3}{n+10}$
  separately
  
  The limit of 5 is obviously just 5 because it is constant and then the limit of the second portion is as such
  
  $\displaystyle \frac{3}{n+10}$
  
  $\displaystyle \frac{\frac{3}{n}}{\frac{n}{n}+\frac{10}{n}}$
  
  which then each indiviual limit is taking your whole limit is equal to
  
  $\displaystyle \frac{0}{1}$ 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, 01:03 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 $\displaystyle x\to\infty$
  $\displaystyle 5-\frac{3}{x+10}$
  
  Similarly by properties of limits we can do
  
  The limit as for $\displaystyle x\to\infty$
  $\displaystyle 5$ - $\displaystyle \frac{3}{n+10}$
  separately
  
  The limit of 5 is obviously just 5 because it is constant and then the limit of the second portion is as such
  
  $\displaystyle \frac{3}{n+10}$
  
  $\displaystyle \frac{\frac{3}{n}}{\frac{n}{n}+\frac{10}{n}}$
  
  which then each indiviual limit is taking your whole limit is equal to
  
  $\displaystyle \frac{0}{1}$ 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 $\displaystyle \epsilon-N$ argument.
  
  We want to show that $\displaystyle \forall~\epsilon>0$, $\displaystyle \exists N$ such that $\displaystyle n>N$ implies
  
  $\displaystyle |L-a_n|=\left|5-\left(5-\frac{3}{n+10}\right)\right|<\epsilon$
  
  $\displaystyle \left|5-\left(5-\frac{3}{n+10}\right)\right|=\frac{3}{n+10}$
  
  To solve $\displaystyle \frac{3}{n+10}<\epsilon$, we get $\displaystyle \frac{n+10}{3}>\frac{1}{\epsilon}\implies n+10>\frac{3}{\epsilon}\implies n>\frac{3}{\epsilon}-10$
  
  Now that the algebra is done, the actual proof is quite simple.
  
  Proof: Given an $\displaystyle \epsilon>0$, let $\displaystyle N=\frac{3}{\epsilon}-10$. So for $\displaystyle n>N$,
  
  $\displaystyle \left|5-\left(5-\frac{3}{n+10}\right)\right|=\frac{3}{n+10}<\frac{ 3}{(\frac{3}{\epsilon}-10)+10}=\frac{3}{3/\epsilon}=\epsilon$
  
  and we're done.
  
  Note: Most professors won't make you show the work to actually find what $\displaystyle N$ should be. My experience has been that you can just assert what $\displaystyle N$ should be, and as long as it works then it's fine.
  • Nov 16th 2009, 02:38 PM
  RockHard
  Hmm it is slowly starting to make some sense, does the order of this statement
  
  $\displaystyle |{a_n - L}| <\epsilon$
  
  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
  
  $\displaystyle |{a_n - L}| <\epsilon$
  
  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, 02:45 PM
  Drexel28
  Quote:
  
  Originally Posted by RockHard
  Hmm it is slowly starting to make some sense, does the order of this statement
  
  $\displaystyle |{a_n - L}| <\epsilon$
  
  matter?
  
  What do you mean here? If you mean does it matter whether it is $\displaystyle \left|a_n-L\right|<\varepsilon$ or $\displaystyle \left|L-a_n\right\|<\varepsilon$ the answer is no. This is because $\displaystyle \left|a_n-L\right|=\left|-\left(L-a_n\right)\right|=|-1|\left|L-a_n\right|=\left|L-a_n\right|$. In other words, absolute values dont care about signs. If you meant does it matter whether it is $\displaystyle \left|a_n-L\right|<\varepsilon$ or $\displaystyle \left|a_n-L\right|>\varepsilon$ the answer is...absolutely. One is saying you can make $\displaystyle a_n$ arbitrarily close...the other is saying, well that you can make their difference arbitrarily large.
  • Nov 16th 2009, 02:47 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, 02:53 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 $\displaystyle \lim_{n\to\infty}a_n=L$ is that just like was said above I can make $\displaystyle a_n$ and $\displaystyle L$ as arbitrarily close (since it's arbitrary let's call that distance epsilon) by making $\displaystyle n$ sufficiently large. In other words if you give me ANY $\displaystyle \varepsilon>0$ (it really does not matter what it is) I need to find some $\displaystyle N\in\mathbb{N}$ such that $\displaystyle n\ge N\implies \left|a_n-L\right|<\varepsilon$. So if a teacher asked you to prove that $\displaystyle \lim_{n\to\infty}\frac{n}{n+1}=1$ what it would mean is that if I give you ANY $\displaystyle \varepsilon>0$ (think of it like a variable...but one that is fixed under specific cases) you can tell me the EXACT $\displaystyle N$ such that $\displaystyle n\ge N\implies \left|\frac{n}{n+1}-1\right|<\varepsilon$. Does that help?
  • Nov 16th 2009, 02:55 PM
  RockHard
  Yes very much so thanks, going for another example to do
  • Nov 16th 2009, 04:01 PM
  RockHard
  Can anyone provide me with another example before I deem this thread solved
  Show 40 post(s) from this thread on one page
  Page 1 of 3 123 Last
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The text extract is a discussion thread from a math forum where users are discussing the concept of limits in sequences. The thread starts with a user named RockHard expressing difficulty in understanding the formal definition of the limit of a sequence. Other users, such as rubix, tonio, and redsoxfan325, provide explanations and examples to help RockHard grasp the concept. The thread includes a detailed example of proving the limit of a sequence using the epsilon-N definition.
  
  The document contains a question and multiple answers to this question.
  
  Label: question-and-answer