L'Hopital's rule proof questions (1 Viewer)

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I've a question concerning spivak's proof of L'Hopital's rule (in chapter 11).

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)
 
Well, correct me if I am mistaken, but it's just needed to prove that such a function c exists, and no more. And the mean value theorem does that, right?
 

HallsofIvy

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I've a question concerning spivak's proof of L'Hopital's rule (in chapter 11).

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'
I feel sure Spivak says nothing of the sort! What is true is that
[tex]\lim_{x\rightarrow a} \frac{f(x)}{g(x)}= \frac{\lim_{x\rightarrow a} f'(x)}{\lim_{x\rightarrow a} g'(x)}[/tex]

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.
c is not necessarily a function of x for the reasons you mention- however, all values of c, if there are more than one, must lie between x and a, for all x, and so as x goes to a, all such values of c must also go to a.

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)
For all x> a, [itex]a\le c\le x[/itex]. Given any [itex]\epsilon> 0[/itex] there exist [itex]\delta> 0[/itex] such that if [itex]|x-a|< \delta[/itex], [itex]|c- a|< |x-a|< \epsilon[/itex].

A similar proof works for x< a.
 
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Well, correct me if I am mistaken, but it's just needed to prove that such a function c exists, and no more. And the mean value theorem does that, right?
Hi I am currently struggeling with the proof of L'hopital anf yhis is one of two concerns for me. Why do we only need to proof that there exists one such formula? I just keep beleving that this only shows something about that function not sure what.

Here is the formula F they use in my book:

http://bildr.no/view/953799

the rest of the proof is here if someone should need it:
http://bildr.no/view/953800

http://bildr.no/view/999144
 

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