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Help with math Sequence

  1. Nov 3, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the limit of the sequence whose terms are given by [tex]a_{n}[/tex]=[tex]n^{2}[/tex](1-cos[tex]\frac{4.2}{n}[/tex])


    2. Relevant equations



    3. The attempt at a solution
    I now that [tex]n^{2}[/tex] goes to infinity so have to use l'hospital rule because you will have infinity*0 which is an indeterminate form, so rewrote as
    [tex]a_{n}[/tex]=[tex]\frac{(1-cos\frac{4.2}{n})}{\frac{1}{n^{2}}}[/tex]

    [tex]lim_{n\rightarrow\infty}[/tex] [tex]\frac{(1-cos\frac{4.2}{n})}{\frac{1}{n^{2}}}[/tex] [tex]\frac{}{}[/tex]
    = [tex]lim_{n\rightarrow\infty}\frac{ [/tex] {[tex]\frac{-4.2sin\frac{4.2}{n}}{n^{2}}[/tex]}/{[tex]\frac{-2}{n^{3}}[/tex]}

    won't the limit still be 0 i really don't understand any of this can anyone please help
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 3, 2007 #2

    Dick

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    Ok, so l'Hopital gives you the form 8.4*n*sin(4.2/n), which is still 0*infinity. Just apply l'Hopital again like you did the first time.
     
  4. Nov 3, 2007 #3
    okay so if i do it again i get
    8.4 [tex]lim_{n\rightarrow\infty}[/tex] [tex]\frac{sin\frac{4.2}{n}}{\frac{1}{n}}[/tex]
    so in the end i will get
    8.4 [tex]lim_{n\rightarrow\infty}[/tex](-4.2cos(4.2/x))

    did i do it right so far
    so is the answer -35.28 i really don't understand how you are suppose to find the limit can someone please help
     
  5. Nov 3, 2007 #4

    Dick

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    Ooops. I made a mistake. Sorry! It's not 8.4*n*sin(4.2/n). It's -2.1*n*sin(4.2/n). Change the 8.4 to -2.1. There are three minuses and the 4.2 is divided by 2. Now it's easy. As n->infinity what does the cos approach. There's a good reason not to give to explicit hints - because I make too many mistakes.
     
  6. Nov 3, 2007 #5
    so as n approaches infinity cos approaches 0 right
    so do you just multiply -2.1 by -4.2 to get the answer
     
  7. Nov 3, 2007 #6
    yeah i got the right answer thanks so much. but i still think that finding limits for sequences are still hard.
     
  8. Nov 3, 2007 #7

    Dick

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    Homework Helper

    Some are, some aren't. Your's is kind of in the middle. Good job, though. Sorry to confuse you.
     
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