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Concept of limit

  1. Apr 14, 2015 #1
    How can it be proved that as lim n tends to infinity, (n2-1)/(n2 + n + 1) tends to 1 ?
     
  2. jcsd
  3. Apr 14, 2015 #2

    SteamKing

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    Have you tried L'Hopital's Rule?
     
  4. Apr 14, 2015 #3

    jbunniii

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    Divide the numerator and denominator by ##n^2## to obtain
    $$\lim_{n \to\infty} \frac{1 - 1/n^2}{1 + 1/n + 1/n^2}$$
    The numerator and denominator both have limit ##1## as ##n \to\infty##, therefore so does the quotient.
     
  5. Apr 14, 2015 #4
    Well, I actually had a fundamental doubt (silly even) for which reason I had posted the question mainly. I am a newbie to the concept of limits. My doubt is as follows. Why should we look to modify the numerator or the denominator or both? Why not just consider that as n tends to infinity the division in question gives us (infinity)/(infinity) which is not defined. This would lead us to the conclusion that the problem is therefore not solvable. Why can't this just be the case. I have actually observed many problems being solved by modifying the numerator and denominator. But it just seems so obvious that (infinity)/(infinity) is the actual result.
     
    Last edited: Apr 14, 2015
  6. Apr 14, 2015 #5
    you are right.
     
  7. Apr 14, 2015 #6
    you can try L'Hopital's Rule. but you must do it twice.
     
  8. Apr 14, 2015 #7

    jbriggs444

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    Because that would be assuming that the limit of the quotient is the quotient of the limits. That principle is almost true. It holds when all three limits exist (and when the limit of the denominator is non-zero). But when the limit of the numerator and denominator both fail to exist, one is left with no prediction for the limit of the quotient. It is an "indeterminate form".
     
  9. Apr 14, 2015 #8

    Mark44

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    You're not really modifying either the numerator or denominator. All that's happening is the factoring of both num. and denom.

    $$\lim_{n \to \infty}\frac{n^2 - 1}{n^2 + n + 1} = \lim_{n \to \infty}\frac{n^2(1 - 1/n^2)}{n^2(1 + 1/n + 1/n^2)} = \lim_{n \to \infty}\frac{n^2}{n^2} \lim_{n \to \infty}\frac{1 - 1/n^2}{1 + 1/n + 1/n^2}$$
    It's legitimate to split a limit of a product into the product of the limits, if the separate limits exist. The first limit all the way to the right exists and is 1. As already stated, the second limit on the right also exists, and also is 1, so the limit on the left exists and is 1.
     
  10. Apr 15, 2015 #9

    jbunniii

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    Consider a very simple example. Let us define a constant sequence: ##x_n = 1## for all positive integers ##n##. Clearly this sequence has limit ##1##. Now let's rewrite the same sequence another way. Most likely you agree that ##n/n = 1## for every ##n##, so I can write ##x_n = n/n##. I have not changed the value of the sequence for any ##n##, so its limit cannot change: it is still ##1##. It doesn't become ##\infty/\infty## just because the numerator and denominator both approach infinity.

    The key observation is that
    $$\lim_{n \to \infty} \frac{a_n}{b_n}$$
    is not generally the same as
    $$\frac{\lim_{n \to\infty}a_n}{\lim_{n \to \infty} b_n}$$
    unless both ##\lim_{n \to\infty}a_n## and ##\lim_{n \to \infty} b_n## exist (and the latter is nonzero).

    Returning to the original problem, the numerator and denominator of
    $$\frac{n^2 - 1}{n^2 + n + 1}$$
    both approach infinity as ##n \to \infty##, so we cannot conclude anything without doing some more work. The point of modifying the fraction, by dividing the numerator and denominator by ##n^2##, is that we end up with
    $$\frac{1 - n^2}{1 + 1/n + 1/n^2}$$
    In this form, the numerator and denominator approach finite limits (##1## in both cases), so now we can apply
    $$\lim_{n\to\infty}\frac{a_n}{b_n} = \frac{\lim_{n \to\infty}a_n}{\lim_{n \to \infty} b_n}$$
    where ##a_n = 1 - n^2## and ##b_n = 1 + 1/n + 1/n^2##.
     
    Last edited: Apr 15, 2015
  11. Apr 15, 2015 #10

    PeroK

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    Another approach. Define:

    ##f(n) =\frac{n^2 - 1}{n^2 + n + 1}##

    Does ##f(n)##, as n increases, eventually get close to 1 and stay there? Whatever definition of "close" you choose?

    If you say "close" is within 0.001, then eventually it gets close (and stays close).

    If you say "close" is within 0.000001, then eventually it gets close (and stays close).

    No matter how you define "close", f(n) eventually gets that close to 1 and stays close.

    So, you might want to say that

    As n increases, f(n) gets arbitrarily close to 1.

    Or, in shorthand:

    ##\lim_{n \to \infty} f(n) = 1##

    Note that there is no concept here of "plugging in" ##n = \infty##. That is meaningless.
     
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