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Sequence n^2 + n -1 / 3n^2 +1

  1. Sep 20, 2004 #1
    Hello all

    For the sequence n^2 + n -1 / 3n^2 +1 I am trying to find

    lim n^2 + n -1/ 3n^2 +1 = 1/3
    n -> 00


    My solution"

    = 1 - 2n^2 - n +2/ 3n^2 +1 = 1- r(sun-n)

    We need to prove that r(sub-n) approaches 2/3.

    Hence 2n^2 - n + 2 < 2n^3
    3n^2 + 1 > 3n^2

    The remainder r sub n is 2n/3. Is this right? How would I find the remainder?

    Also how do I find an N such that n > N and the difference between f(x) and L is less than 1/ 10, 1/100, and 1/1000.

    Any help would be appreciated.

    Thanks
     
  2. jcsd
  3. Sep 20, 2004 #2

    matt grime

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    divide every element in the expression (use brackets!) by n^2, and use the fact that addition and so on are continuous (lim (a_n+b_n) = lima_n + limb_n when all three limits exist, as does: lim(a_n/b_n) = lim(a_n)/lim(b_n) provided b_n doesn't tend to zero
     
  4. Sep 20, 2004 #3

    HallsofIvy

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    I assume that you mean lim (n->00) (n^2 + n -1)/(3n^2 +1) .

    If that is the case then, dividing both numerator and denominator by n^2,
    you get (1+ 1/n- 1/n^2)/(3+ 1/n) which is NOT 1- rn.

    What happens to 1/n and 1/n^2 as n-> 00?
     
  5. Sep 20, 2004 #4

    Pyrrhus

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    Or, you could put it to its function equivalent, and then apply L'Hospital.
     
  6. Sep 20, 2004 #5
    No i do not want to use L' Hopitals Rule. First off:

    (n^2 + n -1)/(3n^2 +1) = 1 - ( 2n^2 - n +2/ 3n^2 +1 ) = 1 - r(sub-n)

    The question tells me to show that r(sub-n) approaches 2/3. Hence we need to find the remainder r(sub-n). I tried finding the remainder by doing this:

    Find an expression which is greater than (2n^2 + n -1).
    Find an expression which is less than (3n^2 +1)

    I got 2n^3 / 3n^2 = 2n/3. Now that I have found the remainder, can anyone verify that this is correct? Also I really need help in finding n such that n > N and the difference between f(x) and L is less than 1/10, 1/1000, 1/10000. I know that after getting the remainder, the difference between f(x) and L cannot be more than 2n/3. But I dont know what n is to satisfy the inequality.

    Any helps would be greatly appreciated.

    Thanks!
     
  7. Sep 20, 2004 #6
    Any help would be greatly appreciated!
     
  8. Sep 20, 2004 #7

    HallsofIvy

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    Since you have completely ignored what help you have been given, I can't think of anything more to say.
     
  9. Sep 21, 2004 #8
    Method of Limits NOT SIMPLY RULES!

    I am sorry HallsofIvy. However I was just wondering if I wanted to find a certain N such that the difference between N and the limit is less than for example 1/10. How would I use the remainder as a tool? I know that the remainder lets say for arguments sake is 2/n, then the difference between the function and the limit cannot exceed 2/n. Hence the main problem is whether I have to guess and check to find the number N, or could I use the remainder as a tool??


    Any help would greatly be appreciated.



    THanks
     
    Last edited: Sep 21, 2004
  10. Sep 21, 2004 #9
    I guess that you have to guess and check?
     
  11. Sep 21, 2004 #10
    For example if we want to find an n such that f(x) - L is less than 1/10, than n = 1. How do i find this? Do i guess and check?
     
  12. Sep 21, 2004 #11
    You may want to do a forum search for delta-epsilon proofs. The way to do it for a limit at infinity is as follows (correct me if I'm wrong).

    For:
    [tex]
    \lim_{x \to \infty} f(x) = L
    [/tex]
    Assuming L is a finite number, you want to find a N such that
    [tex]
    |x| > N(\epsilon) \implies |f(x) - L| < \epsilon
    [/tex]

    Or am I wrong?

    As to how to find it.. just take out a stack of paper buy a few pencils..

    And to your original question.. I just solved the inequality for [itex]n[/itex] and got stuck at: (I'm not very good at math)
    [tex]
    \begin{align*}
    &4n^2\epsilon - 3n + 3\epsilon -4 < 0 \\
    &n > \frac{3+\sqrt{9+16\epsilon-12 \epsilon^2}}{8\epsilon} \\
    & or\\
    &n > \frac{3-\sqrt{9+16\epsilon-12 \epsilon^2}}{8\epsilon}
    \end{align*}
    [/tex]
    One of those SHOULD be your N...

    It's been a long time since I've done the quadratic equation with an inequality. Perhaps I shouldn't have done that? Anyone more learned than myself care to step in? :P
     
    Last edited: Sep 21, 2004
  13. Sep 21, 2004 #12

    HallsofIvy

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    Okay, although that's not quite the question you originally asked!

    First, it should be clear that the the limit of your original rational function is 2/3- you can see that by, as was originally suggested, dividing both numerator and denominator by n2. (n^2 + n -1) / (3n^2 +1)
    equals (1+ 1/n- 1/n2)/(3+ 1/n2). Obviously as n goes to infinity, each of those fractions with n in the denominator goes to 0 and so the whole fraction goes to 1/3. That's not a "rule", that's a perfectly reasonable use of the concept of "limit".

    If you must for some reason use the definition of limit then, instead of writing the fraction as "1- another fraction" you really should write it as
    "1/3- another fraction".

    "Long division" of n^2+ n-1 by 3n^2+ 1 gives 1/3 + (n- 2/3)/(3n^2+ 1).

    You want (n- 2/3)/(3n^2+ 1)< &epsilon; Since 3n^2+ 1 is always positive, we can multiply both sides of the inequality by 3n^2+ 1 to get
    n- 2/3< 3&epsilon;n^2+ &epsilon or 3&epsilon;n2- n+ &epsilon;+ 2/3> 0. For what n is that true. It would be nice if we could factor it but we can, at any rate, solve 3&epsilon;n2- n+ &epsilon;+ 2/3= 0 using the quadratic equation. n= (1+/- &radic;(1- 4&epsilon;(&epsilon+2/3))/2. Messy, but doable! In the example you give, where &epsilon;= 0.1 this becomes n= (1+/- &radic(1- (.4)(.766)))/2.
    I get n= 0.916 and n= .08. Since x has to be a positive integer that just tells us that ANY n will work! For smaller values of &epsilon; we might find larger values of n but there will still be some value that will work.
     
  14. Sep 21, 2004 #13
    Hmm.. I got a different quadratic equation(above, probably wrong?).

    Time to check over my work !

    By the way.. when you get values for n.. does it mean that n > that value?

    Wait.. after a quick google, in your last example, wouldn't n be bounded between the two values from the q. equation?
     
    Last edited: Sep 21, 2004
  15. Sep 21, 2004 #14
    I'm going to try to help you a little bit more
    |(n^2+n-1)/(3n^2+1)-(1/3)|< or = |(3(n^2+n-1)-(3n^2+1))/3(3n^2+1)|<
    or =|(3n-4)/3(3n^2+1)|<|(3n-3)/3(3n^2+1)|< or =|(n-1)/(3n^2+1)|<
    |n/3n^2|< or= |1/3n|=1/3n<error => try to make the error = (1/10^k) =>
    1/3n<1/10^k => n>(10^k)/3.
    I hope this help you contrigrad.
    wisky40
     
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