Excuse me for bumping... I'm just adding new info to the thread.
My teacher gave me a name of a college professor (who taught my math teacher) to talk to about the problem. I called up, and she agreed with my teacher. But then I told her cos(pi/6) was sqrt(3)/2, not 1/2. She commented on how this all was a while ago and she said I was right about the cos(pi/6) thing. So we discuss the problem, and she comes to the same conclusion we've all reached: The limit is to infinity.
I told this to my math teacher, and she looked suprised, and asked me to explain the problem with her. So that's what I'm going to do tomorrow morning, during my study. (No homework to do anyway... ).
Here's what I'm thinking of saying:
1) Mean Value theorem: f'(c) = (f(b) - f(a))/(b-a)
where a and b are points, and c is in between.
2) Definition of a derivative: f'(c) = lim x->a (f(x)-f(a))/(x-a)
Says the same thing as mean value theorem, but less rigorous.
3) Given two functions, f(x) and g(x):
f'(c)/g'(c) == (f(b) - f(a))/(g(b) - g(a))
The (b-a) expression cancels out.
4) Take the limit as b -> a
lim b->a f'(c)/g'(c) == lim b->a ...
5) Assume that f(a) = g(a) = 0
lim b->a f'(c)/g'(c) == (f(b)/g(b))
6) Due to the Squeeze or Sandwich theorem:
lim b->a f'(b)/g'(b) == f(b)/g(b)
7) if f(a) != 0
lim b->a f'(c)/g'(c) == lim b->a (f(b) - f(a))/g(b)
8) So L'Hopital doesn't work in that case.
Anyone see any holes in that?