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Math Help - A question about simplifying lim..

  1. #1
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    A question about simplifying lim..

    My question might sound a bit strange, maybe not, but it is of great importance for me that someone answers it thoroughly...

    Let's say we have the following sequence >>
    a(n) := n / [(n^2) - 4]

    if i were to calculate the lim n --> infinite i would 1 as the result.
    but if i were to divide both the numerator and the denominator with n the sequence would change to a(n) := 1 / [n - (4/n)] and then the lim n --> infinite would be 0. Now the sequence is not convergent 1 but a nullsequence.

    Why is convergent 0 correct and convergent 1 false? As correct answer i was told 0, but why? why not 1? must i always try and divide + simplify?

    2 >> and is there a clear way to simplify before executing the lim? For example, every variable in the numerator must be of power 1 or something?

    I hope my question is clear for you and you can help me. thank you
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    Re: A question about simplifying lim..

    Quote Originally Posted by nappysnake View Post
    My question might sound a bit strange, maybe not, but it is of great importance for me that someone answers it thoroughly...

    Let's say we have the following sequence >>
    a(n) := n / [(n^2) - 4]

    if i were to calculate the lim n --> infinite i would 1 as the result. not true ...
    but if i were to divide both the numerator and the denominator with n the sequence would change to a(n) := 1 / [n - (4/n)] and then the lim n --> infinite would be 0. Now the sequence is not convergent 1 but a nullsequence.

    Why is convergent 0 correct and convergent 1 false? As correct answer i was told 0, but why? why not 1? must i always try and divide + simplify?

    2 >> and is there a clear way to simplify before executing the lim? For example, every variable in the numerator must be of power 1 or something?

    I hope my question is clear for you and you can help me. thank you
    what makes you think the nth term converges to 1 ?

    \lim_{n \to \infty} \frac{n}{n^2 - 4} = 0 , not 1 ... the n^2 term in the denominator is dominant.
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    Re: A question about simplifying lim..

    whoa, wait a second..what do you mean with dominant? I really don't know anything about this.. :|
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    Re: A question about simplifying lim..

    is the denominator dominant because (n^2)-4>n for n>3? how could i specify the dominant term?
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    Re: A question about simplifying lim..

    for an nth term of a sequence in the form of a quotient ... when the degree of the denominator > degree of the numerator, the limit of the sequence as n tends to infinity is zero.

    you should already be familiar with this concept after studying the end-behavior of rational functions in prior algebra / precalculus courses.
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    Re: A question about simplifying lim..

    and if the degree of the numerator > degree of the denominator then the limit with an n that tends to infinity is infinite?

    unfortunately i never had the opportunity to learn, at least from the school or university books up until now i never did anything like that before..
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    Re: A question about simplifying lim..

    Quote Originally Posted by nappysnake View Post
    and if the degree of the numerator > degree of the denominator then the limit with an n that tends to infinity is infinite?

    that would be correct.

    unfortunately i never had the opportunity to learn, at least from the school or university books up until now i never did anything like that before..
    then go learn something more ...

    Pauls Online Notes : Calculus II - Sequences
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    Re: A question about simplifying lim..

    cool, thank you for the link!
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    Re: A question about simplifying lim..

    I've read the info on the link and the calculus I page, it made many things clearer. Thanks!
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