f(x) = [(1 + sin(x))^(1/ sin(x) ? (1 + x)^(1/x)] and 
g(x) = (ln(1 + x))^3

Now, I need the the "LIMIT as X approaches 0" for each "situation":

[f(x)/g(x)], 

[f'(x)/g('x)], 

[f''(x)/g''(x)], 

[f'''(x)/g'''(x)], 

[f''''(x)/g''''(x)], 

Explain why the results do not violate L?Hopital?s Rule.

Dear Lord!

Ha-ha-ha!

Why, he should have included [f'''''(x)/g'''''(x)]?
Or, he should have offered $2.50 for all of those?

-----------
The f'(x) alone would take me two days---if I could get it.
d/dx [u^v] = v*[u^(v-1)]*du/dx +(u^v)*[ln(u)]*dv/du

d/dx [u^v] = ..........+(u^v)[ln(u)]* dv/dx
Not dv/du.
Sorry.

The problem should not be attempted manually - but with the help of a
Computing System - I am using Mathematica 5.1 - but I just wanted a
confirmation of the answers I got.  The third derivative gives a solid
"e/12Log" - but the fourth derivative gets a division by zero error. 
I was hoping someone was proficient with Mathematica or some other
program.

I'm willing to run this through the subset of Maple that's built into
MathCAD, but there is a right parenthesis missing in the equation for
f(x).  If I assume you meant:

  f(x) = (1 + sin(x))^(1/ sin(x)) ? (1 + x)^(1/x)

then in the limit of x -> 0

  f(x)/g(x) = f'(x)/g'(x) = f''(x)/g''(x) = f'''(x)/g'''(x) = exp(1)/12

and

  f''''(x)/g''''(x) = 11*exp(1)/108



In case anyone's curious,
 
  f'''''(x)/g'''''(x) = 103*exp(1)/840

;-)

Looks like l'Hopital's rule is works just fine here.  Remember, the
rule only applies when lims x-> 0 of f(x) and g(x) (or their
derivatives) are both equal to zero or +/-infinity.  This only holds
up to f''(x) and g''(x).  The lim x->0 of f'''(x) is exp(1)/2 while
the lim x-> 0 g'''(x) is 6.