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A quick limits question

  1. Jun 17, 2010 #1
    1. The problem statement, all variables and given/known data
    limit of n/(n+1)^3 as n approaches infinity

    2. Relevant equations

    3. The attempt at a solution

    The degree of the polynomial in the denominator is greater than the degree of the polynomial in the numerator. Is it correct to conclude that the limit of this sequence, as n approaches infinity is zero?

    I appreciate your help
  2. jcsd
  3. Jun 17, 2010 #2


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  4. Jun 17, 2010 #3


    Staff: Mentor

    You can do this by factoring n^3 out of the denominator, to get:
    [tex]\lim_{n \to \infty}\frac{n}{(n + 1)^3} = \lim_{n \to \infty}\frac{n}{n^3(1 + 1/n)^3}[/tex]
  5. Jun 19, 2010 #4
    Hi !

    When you have a problem with a limit.

    It's necessary to think of factorizing, it's the method !

    You factorise numerator, denominator and you simplify !

    Sorry for my english level, i'm new and french !
  6. Jun 19, 2010 #5

    Gib Z

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    That's a good method when you have the ratio of two polynomials and they are of the form 0/0. eg If you had [tex]\lim_{x\to 2}\frac{x^2-4}{x-2}[/tex], then when you put in x=2, it is 0/0, so to get that limit, you could use your method. But with limits as x goes to infinity, you should dividing top and bottom by the highest power of x, as Mark44 did.

    Welcome to PF, and your English is fine :)
  7. Jun 20, 2010 #6
    Another technique, you can verify with calculator the result.

    But it's necessary to demonstrate with the calcul !!!

    Good Bye
  8. Jun 22, 2010 #7
    Another one:
    put t=n+1
    Clearly, t goes to infinity as n goes to infinity,
    so the limit will be :
    lim of (t-1)/t^3 as t goes to infinity
    Devide top and bottom by t:
    lim of (1 - (1/t) )/t^2 as t goes to infinity = (1-0)/infinity=1/infinity=0
  9. Jun 22, 2010 #8
    Another one:

    Use L'Hopital rule.
  10. Jun 22, 2010 #9


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    Of course a calculator can't verify this, only make it plausible, but I don't think you meant that.
  11. Jun 22, 2010 #10


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    One more approach:

    0 < \frac n {(n+1)^3} < \frac n {n^3} = \frac 1 {n^2}

    and your sequence is trapped between 0 and another sequence that converges to zero.
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