L'Hopital's rule proof questions

In summary, the proof of L'Hopital's rule in Spivak's book uses the Cauchy mean value theorem to show that there exists a certain number c in the interval (a;x) such that f(x)g'(c)=g(x)f'(c). The function c is not necessarily a function of x, but all values of c must lie between x and a as x tends to a. This allows us to conclude that lim (as x tends to a) of c = a. The proof only needs to show that such a function c exists, and the mean value theorem does that.
  • #1
mrbean
3
0
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)
 
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  • #2
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?
 
  • #3
mrbean said:
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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  • #4
mrbean said:
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
 

Related to L'Hopital's rule proof questions

What is L'Hopital's rule?

L'Hopital's rule is a mathematical theorem that allows us to find the limit of a function that is in an indeterminate form, such as 0/0 or ∞/∞. It is named after French mathematician Guillaume de l'Hopital.

What are the conditions for using L'Hopital's rule?

The conditions for using L'Hopital's rule are that the limit must be in an indeterminate form, the functions must be differentiable in a neighborhood of the limit point, and the limit of the quotient of the derivatives must exist.

How do you prove L'Hopital's rule?

L'Hopital's rule can be proved using the Cauchy mean value theorem. It involves taking the limit of the difference quotient and using the mean value theorem to show that it is equal to the limit of the quotient of the derivatives.

Can L'Hopital's rule be applied multiple times?

Yes, L'Hopital's rule can be applied multiple times as long as the conditions for using it are still met. This is known as repeated application of L'Hopital's rule.

What are some common mistakes when applying L'Hopital's rule?

Some common mistakes when applying L'Hopital's rule include not checking if the limit is in an indeterminate form, not checking if the derivatives exist, and not using the correct form of the rule (e.g. using the rule for ∞/∞ instead of 0/0).

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