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Trig limit - is this correct?

  1. May 10, 2012 #1
    I'm still trying to figure out how to do limits of trig functions and I would like to know if this is the correct approach. I know the answer is correct, but not sure if that is just a coincidence.

    1. The problem statement, all variables and given/known data

    lim (x -> 0) of (sin 2x) / (sin3x).

    2. Relevant equations



    3. The attempt at a solution

    First I try to get the equation into the form (sin x) / x or x / (sin x)
    so I multiply by

    (x/x)(3/3)(2/2)

    this gives me

    (1/3)(3x/sin3x)(2/1)(sin2x/2x)

    so I get
    (1/3)(1)(2/1)(1) = 2/3

    Is that the correct approach?
     
  2. jcsd
  3. May 10, 2012 #2
    Yes, it is :smile:

    Edit : Another way to solve this would be the use of L'Hospital principle, but this method is more elegant, in my opinion.
     
  4. May 10, 2012 #3
    Thank you very much for the quick reply.

    I'm trying to do calculus as a selfstudy, and the L'Hospital is not mentioned until another 320 pages.
     
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