Results 1 to 8 of 8

Math Help - limits problem

  1. #1
    Junior Member
    Joined
    Feb 2009
    Posts
    39

    limits problem

    compute limit

    lim cos(3x)-1
    x->0 sin(x)sin(x)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2008
    Posts
    461
    Hi!

    Quote Originally Posted by manalive04 View Post
    compute limit

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

    Yours, Rapha
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2009
    Posts
    39
    Quote Originally Posted by Rapha View Post
    Hi!



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

    Yours, Rapha
    how did u do that????
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Nov 2008
    Posts
    461
    Okay, it is

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

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

    Therefor we use L'Hospital

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

    Thus

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

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

    g(x):= sin(x)sin(x)

    g'(x) = 2sin(x)cos(x)

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

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

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

    Any questions?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Feb 2009
    Posts
    39
    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
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,442
    Thanks
    1863
    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, cos(2x)= cos^2(x)- sin^2(x) and sin(2x)= 2sin(x)cos(x)

    Write 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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Nov 2008
    Posts
    461
    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?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    May 2007
    Posts
    237
    Using l'Hospital's Rule

    \lim_{x \to 0} \ \frac{cos(3x)-1}{sin^2(x)}<br />
=\lim_{x \to 0} \ \frac{-3sin(3x)}{2sin(x)cos(x)}<br />
=\lim_{x \to 0} \ \frac{-3sin(3x)}{sin(2x)}<br />
=\lim_{x \to 0} \ \frac{-9cos(3x)}{2cos(2x)}<br />
=-\frac{9}{2}<br />
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Limits problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 3rd 2010, 06:46 AM
  2. Limits Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 1st 2010, 09:24 AM
  3. limits problem..
    Posted in the Calculus Forum
    Replies: 9
    Last Post: March 25th 2009, 07:32 AM
  4. Limits problem
    Posted in the Calculus Forum
    Replies: 7
    Last Post: March 9th 2009, 06:31 AM
  5. limits problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 13th 2008, 03:17 PM

Search Tags


/mathhelpforum @mathhelpforum