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Homework Help: 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

    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

    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

    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

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

    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:


    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

    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

    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

    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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