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Find the limit of the sequence

  1. Apr 12, 2005 #1
    Find the limit of the sequence whose terms are given by [tex]a_n = (n^2)(1-cos(5.2/n))[/tex]

    well as n->inf, cos goes to 1 right? so shouldnt the limit of this sequence be 0?
     
  2. jcsd
  3. Apr 12, 2005 #2

    saltydog

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    Mathematica returns:

    [tex]\mathop \lim\limits_{n\to \infty}n^2[1-Cos(\frac{a}{n})]=\frac{a^2}{2}[/tex]

    I'd like to know how too.
     
  4. Apr 12, 2005 #3

    shmoe

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    But [tex]n^2[/tex] goes to infinity, so your limit is an indeterminate form, infinity*0. Rewrite as:

    [tex]a_n=\frac{1-\cos{\frac{5.2}{n}}}{\frac{1}{n^2}}[/tex]

    and use l'hopital, or you could use the taylor series for cos.
     
  5. Apr 12, 2005 #4

    saltydog

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    Right, just twice. Thanks.
     
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