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Limit of a trigonometric function

  1. Jul 14, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the limit as x approaches 0 of

    f(x) = 3x / sin 5x

    2. Relevant equations

    I might have to use the fact that "the lim of x approaches 0 of sin x/x = 1"

    3. The attempt at a solution

    I dont know how to start to solve the expresion, I would appreciate any help.
    thanks.
     
  2. jcsd
  3. Jul 14, 2009 #2

    Dick

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    Substitute u=5x. So u also approaches 0. Then use lim u/sin(u)=1 as u->0.
     
  4. Jul 14, 2009 #3

    rl.bhat

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    You can rewrite f(x) as
    f(x) = 3/5*5x/sin5x and the find the limit.
     
  5. Jul 15, 2009 #4
    A good explanation of this problem lies in this https://www.physicsforums.com/showthread.php?t=305154".

    I think the best advice is to combine the above two suggestions. Rewrite the limit as
    [tex]\lim_{x\to0} \frac{3x}{\sin 5x} = \lim_{x\to0} \frac{3}{5} \frac{5x}{\sin 5x} = \frac{3}{5}\lim_{x\to0} \frac{5x}{\sin 5x}[/tex]

    Now let u=5x, and follow the example shown in the posted link.
     
    Last edited by a moderator: Apr 24, 2017
  6. Jul 16, 2009 #5
    thanks a lot guys, the explanation you gave in the other post were useful for me, my book didnt explain that way of substitution a lot.


    right now im having a difficult time subtituting this:

    1.) f(x) 3x^2 / 1 - cos 1/2 x^2

    I think that for this problem I gotta use the fact that x - cos x / x = 0


    and
    2.) g(x) sin x /x^2


    at first sight, these problems arent substutited in the same was as you described in the previous post though. Sometimes when I try to substitute terms I get all confused and end up mading up answers for the trigonometric functions, I was wondering if there are certain substitution steps that I could use for most of the trigonometric functions? like the one you described in the previous posts.

    take care and thanks a lot.
     
  7. Jul 16, 2009 #6

    rock.freak667

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    is this supposed to be

    [tex]\lim_{x \rightarrow 0} \frac{3x^2}{1-cos\frac{x^2}{2}}[/tex]


    [tex]\lim_{x \rightarrow 0} \frac{sinx}{x^2} [/tex]


    if that is the question then try using the fact that 1/x2 = (1/x)*(1/x)
    and

    [tex]\lim_{x \rightarrow a} f(x)g(x) = \lim_{x \rightarrow a}f(x) * \lim_{x \rightarrow a} g(x)[/tex]
     
  8. Jul 17, 2009 #7
    the second problem is right, the first one is 3x^2 / (1-cos 1/2 x)^2
    sorry im having problems with the notation.
     
  9. Jul 17, 2009 #8

    HallsofIvy

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    Then both multiply and divide the numerator by 1/4 to get
    [tex]\frac{12\frac{x^2}{4}}{\left(1- cos \frac{x}{2}\right)^2}= 12\left(\frac{\frac{x}{2}}{1- cos(\frac{x}{2})}\right)^2[/itex]
    and use the fact that
    [tex]\lim_{a\rightarrow 0}\frac{1- cos(u)}{u}= 0[/itex]
    But note that your fraction is the reciprocal of that- that's very important here!
     
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