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Limit question

  1. Mar 10, 2009 #1
    1. The problem statement, all variables and given/known data
    This should be easy but I can't find why.

    Why is the following true for 0<x<1,

    [tex]\lim_{n \rightarrow \infty}nx(1-x^2)^n = 0 [/tex]


    2. Relevant equations



    3. The attempt at a solution

    I understand why (1-x^2)^n goes to zero, but the nx part is not bounded and seems to be going to infinity?
     
  2. jcsd
  3. Mar 10, 2009 #2

    lanedance

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    Homework Helper

    can you use L'hopitals rule here?
     
  4. Mar 10, 2009 #3
    I am not sure. I don't have much experience with L'Hospital's Rule so I will look it up.
     
  5. Mar 10, 2009 #4

    Mark44

    Staff: Mentor

    Not as the limit currently is written. L'Hopital's rule applies to quotients only, where both the numerator and denominator are approaching zero, or where both are approaching +/- infinity. To get this expression into a suitable form, instead of multiplying by n, you can divide by 1/n.
     
  6. Mar 10, 2009 #5

    lanedance

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    Homework Helper

    another possibility for these types of limits is to write:

    [tex]\lim_{n \rightarrow \infty} nx(1-x^2)^n = \lim_{n \rightarrow \infty} e^{\ln{nx(1-x^2)^n}} [/tex]
    and use the log laws to manipulate the expression, will have a go & see how I go
     
    Last edited: Mar 10, 2009
  7. Mar 10, 2009 #6

    lanedance

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    hmm... the only way I manged this limit was to factor

    [tex]
    \lim_{n \rightarrow \infty} nx(1-x^2)^n = \lim_{n \rightarrow \infty} (nx(1+x)^n)((1-x)^n)
    [/tex]

    then manipulating to make use of L'hopitals rule (for 0<x<1)
    [tex]
    = \lim_{n \rightarrow \infty} \frac{(1-x)^n}{\frac{1}{nx(1+x)^n}} \rightarrow \frac{0}{0}
    [/tex]

    and differentiating numerator & denominator w.r.t. n gives the required limit, seems a little tricky though, not sure if I'm missing something...
     
  8. Mar 11, 2009 #7

    Mark44

    Staff: Mentor

    I don't think factoring the 1 - x^2 part makes life simpler. I would rewrite the original limit expression as [tex]x \lim_{n \rightarrow \infty} \frac{(1 - x^2)^n}{1/n}[/tex], and use L'Hopital's Rule on that. Both the numerator and denominator are approaching 0, so L'Hopital's Rule applies.
     
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