L'Hopital's Theorem: If limf(x)=limg(x)=infty, then lim(f/g)

[marlen19861]

I need to prove this version of L'Hopitals theorem.

If limf(x)=limg(x)=infinity (as x goes to infinity) then lim[f(x)/g(x)]=lim[f'(x)/g'(x)] (as x goes to infinity)

Any help please?

Thank you

 

[mmm4444bot]

Re: L'Hopital



This is not an elementary exercise, so I cannot help but assume that you possess some background information.

Did somebody provide a you with a proof for the version which does not involve infinity, and you are now required to prove the version that you posted?

You did not provide any information about why you are unable to begin anything at all toward writing a proof, so I have no idea about what type of help you expect or need. Therefore, I offer the following conjecture.

I speculate that various proofs of all versions of L'Hôpital's Rule are posted on the Internet in such a way as to be easily located through supplication to the Great God Google (GGG).

Cheers,

~ Mark

PS: Please read the post titled "Read Before Posting", if you have not already done so.

 

[daon]

Rudin's proof is very nice.

 

[marlen19861]

Rudin's proof is very nice.

Rudin?

 

[PAULK]

Rudin's proof is very nice.

Rudin?

Elementary Analysis, 73rd Edition, or something. But if you have to work one up yourself, my guess is to use the MVT in some way.

 

[pka]

Here are some very good links.
http://mathforum.org/library/drmath/view/53750.html

http://mathforum.org/library/drmath/view/65515.html

http://mathforum.org/library/drmath/view/53665.html