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Quick question about the range of a sine function

  1. Sep 26, 2009 #1
    1. The problem statement, all variables and given/known data
    I am doing a proof about limits, and I need to know the range that sin(n^3-2n)/n is within for all n.


    2. Relevant equations



    3. The attempt at a solution

    I know that sin(n) is in between -1 and 1 for all n, so sin(n)/n would be in between -1/n and 1/n for all n. However, I don't know how to manipulate the range if the inside of the sin(n) function is changed. Any help would be great!!!
     
  2. jcsd
  3. Sep 26, 2009 #2

    lurflurf

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    Homework Helper

    Is this for n->infinity?
    if so
    sin(n^3-2n) is between -1 and 1
    is what you need
    then consider 1/n
     
  4. Sep 26, 2009 #3
    yes this is for n as n approaches infinity. Would it be between -(n^2-2)/(n^3-2n) and (n^2-2)/(n^3-2n)?
     
  5. Sep 26, 2009 #4
    nevermind i misread your comment, i see that it is in between -1/n and 1/n. Thanks for the help!
     
  6. Sep 26, 2009 #5
    I would find the limit as n->0 and define the function at n=0 to make it continuous. Then find its local maxes and mins from its derivative.
     
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