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, let's 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.
