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Limit of a sequence

  1. Jan 18, 2017 #1
    1. The problem statement, all variables and given/known data

    Suppose [tex] a_{n}=\frac{n^2-2n+1}{2n^2+4n-1} [/tex]

    For each positive number [tex] \epsilon [/tex], find a number N such that:

    [tex] \mid a_{n} - L\mid < \epsilon [/tex] whenever n > N.

    2. Relevant equations


    3. The attempt at a solution

    [tex] \mid \frac{n^2 -2n + 1} {2n^2+4n-1} - \frac{1} {2} \mid < \epsilon [/tex]

    [tex] \mid \frac{3-8n} {4n^2+8n-2} \mid < \epsilon [/tex]

    Now I have no idea how to isolate the n so that I can find the N value. Any help appreciated.

    Thanks!
     
    Last edited: Jan 18, 2017
  2. jcsd
  3. Jan 18, 2017 #2

    PeroK

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    Try some estimating.

    Can you fix your latex?
     
  4. Jan 18, 2017 #3
    Fixed it up.

    Estimating what exactly?
     
  5. Jan 18, 2017 #4

    PeroK

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    Estimating is a technique, which no one seems to teach explicitly for evaluating limits. Let me give you an example and see if you can apply the idea to your case.

    Show that ##5n^2 + 3n + 133## tends to ##\infty## an ##n \rightarrow \infty##

    Estimating technique:

    For ##n > 1## we have ##n^2 > n## hence ##5n^2 > n##. So:

    ##5n^2 + 3n + 133 > 4n + 133 > 4n > n##

    Now, we can use a simple ##N## in the limit argument. Whereas, trying to find ##N## for the orginal quadratic would be messy.
     
  6. Jan 18, 2017 #5

    PeroK

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    PS I think you've lost a factor of ##2## from the denominantor.
     
  7. Jan 18, 2017 #6
    Yes I definitely have, thanks.
     
  8. Jan 18, 2017 #7
    So it goes kinda likes this:

    For n > 1 we have [tex] n^2>n [/tex] hence, [tex] 2n^2>n [/tex] So:

    [tex] 2n^2+4n-1>5n-1>5n>n [/tex]

    And in a similar way:

    For n > 1 we have [tex] 3 - 8n > 3 - 8 = -5 [/tex] So:

    [tex] \mid \frac{-5}{n} \mid < \epsilon [/tex]

    and,

    [tex] n > 5 / \epsilon [/tex]

    Is that what you mean?
     
  9. Jan 18, 2017 #8

    PeroK

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    First, you need to take that ##\epsilon## out of your working. That seems to be a common mistake to introduce ##\epsilon## into the inequality too soon. Second, your estimates have gone a bit wonky!

    Let me show you what I mean. You should leave out the ##\epsilon## for now and do:

    ##\mid \frac{n^2 -2n + 1} {2n^2+4n-1} - \frac{1} {2} \mid \ = \ \mid \frac{3-8n} {4n^2+8n-2} \mid \ < \ \mid \frac{8n} {4n^2+8n-2} \mid \ \dots##

    Note that the numerator is less than ##8n## and you need an over-estimate on the numerator and an underestimate on the denominator. But, your underestimate on the denominator can't be too crude, because ##8n/n## doesn't converge to 0. So, you need an estimate based on ##n^2## in this case.

    It's only when you've finished this algebra that you bring in ##\epsilon##. For example:

    Let ##\epsilon > 0 ## and take ##N > 1/ \epsilon## (or whatever you think you need from your calculations). Then ##n > N## implies:

    ##\mid \frac{n^2 -2n + 1} {2n^2+4n-1} - \frac{1}{2} \mid < \frac{1}{n} < \epsilon##
     
  10. Jan 18, 2017 #9
    Ok I think I get it now. Thanks for your help I'll do some more work on it.
     
  11. Jan 18, 2017 #10

    PeroK

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    The simplest way to explain estimating is perhaps:

    ##|\frac{a(n)}{b(n)}| < |\frac{c(n)}{d(n)}|##

    Where ##|a(n)| < |c(n)|## and ##|b(n)| > |d(n)|## and ##c(n), d(n)## are some simple expressions in ##n## that reduce to something that can easily be shown to have a limit of ##0##.
     
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