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Squeeze theorem question

  1. Apr 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Find lim sin4x/3x
    x -> 0


    2. Relevant equations



    3. The attempt at a solution
    I did some algebraic massaging and got

    sin4/3x = 4/4 * sin4x/3x = 4/3 * sin4x/4x
    but I don't understand how sin4x/4x becomes 1 like my text book says so.
     
  2. jcsd
  3. Apr 9, 2012 #2

    Dick

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    sin(4x)/(4x) isn't 1. The limit as x->0 of sin(4x)/(4x) is 1. Have you proved limit u->0 of sin(u)/u=1? Then just put 4x=u.
     
    Last edited: Apr 9, 2012
  4. Apr 9, 2012 #3
    Yeah you're right.So, It's true that limit u->0 of sin(u)/u=1?
     
  5. Apr 9, 2012 #4

    Dick

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    Didn't you prove limit u->0 sin(u)/u=1. I doubt they would have given you this problem if they hadn't.
     
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