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Thread: limits problem

  1. #1
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    limits problem

    compute limit

    lim cos(3x)-1
    x->0 sin(x)sin(x)
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  2. #2
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    Hi!

    Quote Originally Posted by manalive04 View Post
    compute limit

    lim cos(3x)-1
    x->0 sin(x)sin(x)
    $\displaystyle \lim_{x \to 0} \ \frac{cos(3x)-1}{sin(x)sin(x)} = - \frac{9}{2}$

    Yours, Rapha
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  3. #3
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    Quote Originally Posted by Rapha View Post
    Hi!



    $\displaystyle \lim_{x \to 0} \ \frac{cos(3x)-1}{sin(x)sin(x)} = - \frac{9}{2}$

    Yours, Rapha
    how did u do that????
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  4. #4
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    Okay, it is

    $\displaystyle \lim_{x \to 0} \ cos(3x)-1 = 1-1 = 0$

    $\displaystyle \lim_{x \to 0} \ sin(x)sin(x) = 0$

    Therefor we use L'Hospital

    We know that $\displaystyle cos(3x) = 4cos(x)^3 - 3cos(x)$

    Thus

    $\displaystyle cos(3x)-1 = 4cos(x)^3 - 3cos(x) =:f(x) $

    => $\displaystyle f'(x) = - 8sin(x)*cos(x)^2 + 4*sin(x)^3 - sin(x)$

    $\displaystyle g(x):= sin(x)sin(x)$

    $\displaystyle g'(x) = 2sin(x)cos(x)$

    => $\displaystyle \lim_{x \to 0} \ \frac{cos(3x)-1}{sin(x)sin(x)}$

    = $\displaystyle lim \ \frac{- 8sin(x)*cos(x)^2 + 4*sin(x)^3 - sin(x)}{2sin(x)cos(x)}$

    = $\displaystyle lim \ \frac{-8cos^2(x)+4sin^2(x)-1}{2cos(x)} = \frac{-8+4*0-1}{2*1} = -9/2$

    Any questions?
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  5. #5
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    how did you calculate the cos(3x) how is that broken down?

    i know cos(2x) is cos(x)cos(x) - sin(x)sin(x)

    but how did u do this one
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  6. #6
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    Quote Originally Posted by manalive04 View Post
    how did you calculate the cos(3x) how is that broken down?

    i know cos(2x) is cos(x)cos(x) - sin(x)sin(x)

    but how did u do this one
    Yes, $\displaystyle cos(2x)= cos^2(x)- sin^2(x)$ and $\displaystyle sin(2x)= 2sin(x)cos(x)$

    Write $\displaystyle cos(3x)= cos(2x+ x)$ and use the sum formula: cos(a+ b)= cos(a)cos(b)- sin(a)sin(b) so that cos(3x)= cos(2x)cos(x)- sin(2x)sin(x). Now put the formulas form cos(2x) and sin(2x) into that.
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  7. #7
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    Hi again.

    Quote Originally Posted by manalive04 View Post
    how did you calculate the cos(3x) how is that broken down?

    i know cos(2x) is cos(x)cos(x) - sin(x)sin(x)

    but how did u do this one

    I'm glad you ask, but what exactly is the question?

    Click here:

    List of trigonometric identities - Wikipedia, the free encyclopedia

    Use this formula for n=3


    ???

    Does it help?
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  8. #8
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    Using l'Hospital's Rule

    $\displaystyle \lim_{x \to 0} \ \frac{cos(3x)-1}{sin^2(x)}
    =\lim_{x \to 0} \ \frac{-3sin(3x)}{2sin(x)cos(x)}
    =\lim_{x \to 0} \ \frac{-3sin(3x)}{sin(2x)}
    =\lim_{x \to 0} \ \frac{-9cos(3x)}{2cos(2x)}
    =-\frac{9}{2}
    $
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