- #1

mrbean

- 3

- 0

It goes like this,

lim (x tends to a) of f = 0

lim (x tends to a) of g = 0

lim (x tends to a) of f'/g' exists,

then,

lim (x tends to a) f/g exists and is equal to lim (x tends to a) of f'/g'

He clarifies some thing that are implicit in the hypothesis, shows that g(x) is not equal to zero in some interval containing a, and then,

He applies Cauchy's mean value theorem to say there exists a certain number c (which is indexed to x) in (a ; x) such that f(x)g'(c) = g(x)f'(c). Then he says that as x tends to a,

c tends to a. And after some other step, he finishes.

The whole idea of the proof is clear, but:

My questions here are:

1. As c is indexed to x, is c a function of x? If so, can you give a proof of this? ( My problem here is that when I define the function c over a certain interval containing a, such that c(x) satisfies,

f(x)g'(c(x)) = g(x)f'(c(x)), Cauchy's mean value theorem assures existence of at least one c(x), but what if there are more? Then c wouldn't be a function, wouldn't it? How do you fix this? I know this is obvious, and that you can fix it, but I can't see how to formalize it.

2. Can we then say then that lim (as x tends to a) of c = a ? ( If the above question is answered, this is easy, but otherwise I don't see how to prove it)