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I have a midterm test 2moro and i need help

  1. Feb 10, 2008 #1
    1. The problem statement, all variables and given/known data
    lim x cos x-sin x/x sin^2 x
    x-0



    2. Relevant equations



    3. The attempt at a solution
    ...............................................................
    no clue
    i just know that we should use l'hopitals rule
     
  2. jcsd
  3. Feb 10, 2008 #2
    L'Hopital's rule states that if you have a limit of a fraction where both the numerator and denominator of the fraction go to 0 or infinity (in this case infinity), the limit is equal to the same limit of the derivatives of the numerator and denominator.

    So in your particular limit, lets assume that f(x)=x*cosx-sinx and g(x)=s*sin^2 x

    Then first you must show that both the numerator and denominator go to 0 as x goes to 0.

    lim x*cosx-sinx
    x->0

    This goes to 0 because as x goes to 0, x*cosx goes to 0, and because sin(0) is 0, so does sinx.

    lim x*sin^2 x
    x->0

    This also goes to 0 because sin goes to 0 as well as x.

    So because of this, L'Hopital's rule then states:

    lim x*cosx-sinx/x*sin^2 x = lim f(x)/g(x) = lim f'(x)/g'(x) (all of these limits are as x->0)


    so we then take the derivatives of f(x) and g(x).

    Using the product rule we find that:
    f'(x) = cosx-x*sinx-cosx = -x*sinx

    Using the same rule we find that:
    g'(x) = 2*x*cosx*sinx+sin^2 x

    So the limit is now:
    lim -x*sinx/2*x*cosx*sinx+sin^2 x
    x->0

    canceling sinx makes:
    lim -x/2*x*cosx+sinx
    x->0

    Now if you look at this new limit, you can see that the top and bottom once again both go to 0. So L'Hopital's rule can be applied once again:

    f(x)=-x
    g(x)=2*x*cosx+sinx

    it is easily seen that f'(x)=-1

    From the product and addition rule we see that:
    g'(x)=2*cosx-2*x*sinx+cosx = 3*cosx-2*x*sinx

    So our new (and final) limit is:

    lim -1/3*cosx-2*x*sinx
    x->0

    Now, in this limit, we only need to break up the terms.

    2*x*sinx will go to 0, so we can cancel that out, however, 1, and 3*cosx do not.

    cos(0)=1, so 3*cosx as x->0 is equal to 3.

    So in the end we have the following limit:

    lim -1/3*cosx
    x->0

    And because 1 goes to 1, and 3*cosx goes to 3, we obtain our final answer:

    lim x*cosx-sinx/x*sin^2 x = -1/3
    x->0

    Sorry that I didn't use LaTex for the equations. So if they look a little unclear in their format, writing them down will help.
     
    Last edited: Feb 10, 2008
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