Results 1 to 4 of 4

Math Help - L'Hospital's Rule

  1. #1
    Member
    Joined
    Nov 2009
    Posts
    92

    L'Hospital's Rule

    Use L'hospital's Rule to evaluate

    Lim x -->0 9x(cos5x-1)/sin(2x-2x)

    I know that this is a 0/0 indeterminate form and you need to differentiable both top and bottom until you can substitute 0 into x. I have problems differentiating sin(2x-2x).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Dec 2008
    Posts
    319
    Do you mean \sin 2x - 2x? If so, we may differentiate it as follows:

    \frac{d}{dx}(\sin 2x - 2x)=\frac{d}{dx}(\sin 2x)-\frac{d}{dx}(2x)=2\cos 2x-2.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2009
    Posts
    92
    Yes, thats write, so how to do compute the limit afterwards? i got something like -45x*sin5x+9cos5x-9/2cos2x-2
    This would still result in [0,0]. So i would differentiate once again.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Nov 2009
    Posts
    92
    my final answer i got is 675/8, I want to check if this is correct. Thanks to everyone!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. without using L'Hospital's Rule
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 31st 2010, 02:48 AM
  2. L'Hospital's rule
    Posted in the Calculus Forum
    Replies: 8
    Last Post: February 16th 2010, 08:37 PM
  3. L'Hospital's Rule
    Posted in the Calculus Forum
    Replies: 10
    Last Post: December 9th 2009, 04:26 PM
  4. L'Hospital's rule
    Posted in the Calculus Forum
    Replies: 10
    Last Post: November 19th 2009, 08:31 PM
  5. L'Hospital's Rule
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 15th 2008, 08:24 PM

Search Tags


/mathhelpforum @mathhelpforum