From: Bill Dubuque <wgd@berne.ai.mit.edu>
Subject: Re: a help with a limit
Date: 23 Jan 1999 15:57:09 -0500
Newsgroups: sci.math
Keywords: Fun with L'Hospital's Rule

acst@my-dejanews.com wrote:
| Can we always guarantee that, if f is a function from R to R, differentiable
| on R and such that  lim f(x) = k as x->oo  then  lim f'(x) = 0 as x->oo ?
| If it's always true, can anyone give me a rigorous proof?

As others have already pointed out, there are easy counterexamples 
if lim f' does not exist. When lim f' exists the result holds true;
in fact there is an elegant proof via L'Hopital's rule of the
following more general result (where L may be infinite):

                                   f e^x       (f + f') e^x
 lim  f + f' = L  =>  lim f = lim  ----- = lim ------------ = L
x->oo                x->oo   x->oo  e^x   x->oo     e^x

Thus, in your case, when lim f = L is finite and lim f' exists
it follows that lim f' = 0. (There are obvious generalizations;
e.g. replace e^x by appropriate g(x) with infinite limit,
e.g. with g(x) = log_n(x) = log(log(...(x)...))  [n logs]

 lim f(x) = lim(f(x) + f'(x) x log(x) log(log(x)) ... log_n(x))

if the latter limit exists).

The above result goes back to Hardy's classic calculus text
(A Course of Pure Mathematics, p. 281 (50)), although his 
proof is more awkward than the elegant proof above using
L'Hospital's rule - which is folklore and often rediscovered, 
e.g. see [1], which has some history and generalizations.

[1] Martin D. Landau; William R. Jones.  A Hardy Old Problem.
Math. Magazine 56 (1983) 230-232.

-Bill Dubuque

==============================================================================

From: hwatheod@leland.Stanford.EDU (theodore hwa)
Subject: Re: an example about Hopital's rule
Date: 18 Nov 1999 18:02:14 GMT
Newsgroups: sci.math

Marco Marzocchi (mrz@dmf.bs.unicatt.it) wrote:
: Hi,
: 
: once I found in a textbook an example about Hopital's
: rule of the following kind:
: 
: applying twice Hopital's rule to f/g one obtains back
: f/g itself (i don't remember if there was also some
: constant multiplying it), so the rule itself cannot be
: applied to get the result.
: 
: Does anybody remembers it, or remember the title of the
: book where it appears?

This is tantamount to solving f'' = cf, whose solutions are combinations
of exponentials.  You always want limits to go to 0 or infinity, to make
L'hopital's rule applicable.  So try something like

lim x->inf (e^(2x) + e^(-2x))/(e^(3x)+e^(-3x))

or for something really simple

lim x->inf e^x / e^x

:)

Ted

==============================================================================

From: "Charles H. Giffen" <chg4k@virginia.edu>
Subject: Re: an example about Hopital's rule
Date: Tue, 30 Nov 1999 12:36:09 -0500
Newsgroups: sci.math
To: Marco Marzocchi <mrz@dmf.bs.unicatt.it>

Marco Marzocchi wrote:
[as above -- djr]

Let  f(x) = x  and  g(x) = sqrt(1+x^2).  Consider
the limit of  f(x)/g(x)  as  x --> infinity.  Since
f(x) --> infinity  and  g(x) --> infinity, l'Hospital's
rule may be tried, yielding

f'(x)/g'(x) = 1/(x/sqrt(1+x^2)) = sqrt(1+x^2)/x = g(x)/f(x).

Applying l'Hospital's rule now to  g(x)/f(x)  returns  
g'(x)/f('x) = f(x)/g(x), which puts back where we started.

--Chuck Giffen