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Limit of trigonometry

  1. Apr 29, 2013 #1
    1. The problem statement, all variables and given/known data
    [tex]\lim_{x \to 0} \frac{tan (cos 4x - 1)}{3x ~ sin (\frac{4}{3} x)}[/tex]


    2. Relevant equations
    limit for trigonometry


    3. The attempt at a solution
    can I do it like this:

    [tex]\frac{tan (cos 4x - 1)}{3x ~ sin (\frac{4}{3} x)}[/tex]

    [tex]= \frac{- tan (2 sin^{2} 2x)}{3x ~ sin (\frac{4}{3} x)}[/tex]

    and then using the property of trigonometry limit, it becomes:

    [tex]= \frac{-2 . 4}{3 . \frac{4}{3}}[/tex]

    [tex]=-2[/tex]
     
  2. jcsd
  3. Apr 29, 2013 #2
    What do you know about sin(x), cos(x) and tan(x) when x is very small?
     
  4. Apr 29, 2013 #3
    I am not sure what you mean, maybe like this:

    a. when x is very small (close to zero):
    the value of sin x is close to 0
    the value of cos x is close to 1
    the value of tan x is close to 0

    or

    b. when x is very small (close to zero):
    sin x ≈ x
    cos x ≈ 1 - 1/2 x2 ≈ 1
    tan x ≈ x

    but I still don't know what the properties related to the question
     
  5. Apr 29, 2013 #4

    Mentallic

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    Homework Helper

    You're looking to use the properties of b.

    If [itex]\cos(x)\approx 1-x^2/2[/itex] then what is [itex]\cos(4x)[/itex] approximately equal to?

    What's [itex]\sin(4x/3)[/itex] approximately equal to?

    Finally, you'll need to also convert the tan function as well in the same fashion.
     
  6. May 2, 2013 #5
    Oh I see. I don't know before that the properties can be used in limit as well.

    Thanks a lot for all the help
     
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