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Find the Limit

  1. Feb 28, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the limit
    lim (sin(3x))/x
    x-->0


    2. Relevant equations
    There is two steps which I don't understand how they work, or why they work. Ill show you.
    I'm using my answer book to help me out, but it doesn't give a satisfactory explanation.


    3. The attempt at a solution
    lim (sin(3x))/x
    x-->0

    = lim 3(sin(3x))/3x
    x-->0

    = 3lim (sin(3x))/3x as x -> 0 3x -> 0 Why do I do this maneuver? I understand its for
    x-->0 substitution

    = 3lim (sin(3x))/3x let theta = 3x

    = 3lim (sin(theta))/theta

    = 3(1) Why does lim (sin(theta))/theta equal 1?
     
  2. jcsd
  3. Feb 28, 2014 #2

    hilbert2

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    Gold Member

    Here's a geometrical argument that 'proves' the limit is equal to 1: http://math.ucsd.edu/~wgarner/math20a/sin%28x%29_over_x.htm [Broken] .

    I think this argument is enough for your proof, you're probably not supposed to deduce the limit from the Maclaurin series expansion of the sine function (which would be the rigorous way to do it).
     
    Last edited by a moderator: May 6, 2017
  4. Feb 28, 2014 #3

    ChrisVer

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    Gold Member

    There are many ways to check it... first of all you have a form of 0/0 for which you can use the del'hopitale...
    Another way is the Taylor expansion of sine : the argument is small so you just need the first order term which is [itex]\theta[/itex] and gives 1 with the denominator...
    Another way is via the sandwich rule demonstrated by hilbert12's reference.
     
  5. Feb 28, 2014 #4
    Thank you very much. I like to learn how it works for some odd reason, ocd or something. Its like i doubt maths even though its been around for ever lol. Thanks.
     
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