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Limit of sequence lim n (1-cos(2/n))

  1. Feb 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Untitled.jpg

    2. Relevant equations



    3. The attempt at a solution

    above picture !!

    Exam in 3 hours :( please help
     
  2. jcsd
  3. Feb 21, 2013 #2

    jbunniii

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    You can write
    $$\begin{align}
    1 - \cos\left(\frac{2}{n}\right) = 2\left[\frac{1}{2} - \frac{1}{2}\cos\left(\frac{2}{n}\right)\right]
    &= 2\sin^2\left(\frac{1}{n}\right) \\
    \end{align}$$
     
  4. Feb 21, 2013 #3

    LCKurtz

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    $$\frac 1 {\frac 1 n}$$doesn't go to 0 as ##n\to\infty##.

    Another method is to use L'Hospital's rule on$$
    \frac{1 - \cos{\frac 1 n}}{\frac 1 n}$$
     
  5. Feb 21, 2013 #4
    where is 'n' gone?
     
  6. Feb 21, 2013 #5
    thanks LCKurtz
     
  7. Feb 21, 2013 #6

    jbunniii

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    It didn't go anywhere. I was just suggesting how to rewrite the hard part. Now if we substitute back, we get
    $$n\left[1 - \cos\left(\frac{2}{n}\right)\right] = 2n \sin^2\left(\frac{1}{n}\right)$$
    Can you see what to do now?
     
  8. Feb 21, 2013 #7

    Thanks jbunniii
     
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