l'hospital's rule - why the condition?
Luke Wu
lim(x->a) f(x) / g(x)
Why is there a condition that g(x) <> 0 when x is in an open interval
that contains a, except possibly at a before we can apply l'hospitals
rule (assuming it's a 0/0 form)?
Does anyone have an intuitive explanation (on top of a rigorous) for
this condition?
Luke Wu
lim(x->a) f(x) / g(x)
Why is there a condition that g(x) <> 0 when x is in an open interval
that contains a, except possibly at a?
Jules
If this condition were not satisfied, then f(x) / g(x) would not be
defined in a punctured neighborhood of a, so the limit would not exist.
Jonas
"Luke Wu" <LookSk...@gmail.com> wrote in message
news:1136680800.5...@o13g2000cwo.googlegroups.com...
because otherwise lim(x->a) g(x)=0 and you would have division by zero.
David C. Ullrich
wrote:
No, actually lim(x->a) g(x)=0 is one of the _hypotheses_ in one
version of the rule. Having lim(x->a) g(x)=0 is not a bad
thing here. But we have to have g <> 0 in some deleted neighborhood
of a, otherwise we _do_ have division by zero.
>
************************
David C. Ullrich
The World Wide Wade
<1136680800.5...@o13g2000cwo.googlegroups.com>,
"Luke Wu" <LookSk...@gmail.com> wrote:
> In the limit
>
> lim(x->a) f(x) / g(x)
>
> Why is there a condition that g(x) <> 0 when x is in an open interval
> that contains a, except possibly at a before we can apply l'hospitals
> rule (assuming it's a 0/0 form)?
There is no good reason for this assumption, as it follows from
the other assumptions in LHR. We are assuming f'/g' has a limit
at a. But that can't even get off the ground unless g' is nonzero
in a deleted neighborhood of a. We also assume g(x) -> 0 at a; if
we define g(a) = 0, then g is continuous at a. The MVT then gives
g(x) = g(x) - g(a) = g'(c)(x-a), which is nonzero for x close to
(but not equal to) a.
Zdislav V. Kovarik
Intuitive or otherwise:
Go through the proof of L'Hospital's Rule and count how many times it used
-> Rolle's Theorem, or
-> Mean Value Theorem, or
-> Daboux's Theorem (IVT for derivatives), or
-> Intermediate Value theorem
(depending on the taste of the person who presents a proof).
Each of the theorems assumes that a certain function is defined (and/or
differentiable) on an interval, without any "holes".
(Two-sided limits are treated by separate arguments for left and for
right limit, to work around the required "hole" at the point a.)
If the domain has too may "holes" near the point a, preventing the use of
those theorems, the proof will not work.
Even after we relax the definition of a limit to allow for holes in the
domain, as long as the point a is an accumulation point,
I suspect that a counterexample can be constructed, showing that the
assumption you quoted cannot be thrown away,
Counterexample anyone?
Cheers, ZVK(Slavek).
Robert Israel
> Counterexample anyone?
I don't think so.
Suppose f and g are differentiable in a neighbourhood of a,
lim_{x -> a} f(x) = lim_{x -> a} g(x) = 0, and a is an
accumulation point of {x: g(x) = 0}. By Rolle's Theorem, between
any two zeros of g there is a zero of g'. Since f'(x)/g'(x)
is undefined when g'(x) = 0, lim_{x -> a} f'(x)/g'(x) does
not exist. So this can't be a counterexample.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Dave L. Renfro
> If the domain has too may "holes" near the point a,
> preventing the use of those theorems, the proof
> will not work.
>
> Even after we relax the definition of a limit to
> allow for holes in the domain, as long as the point
> a is an accumulation point,
>
> I suspect that a counterexample can be constructed,
> showing that the assumption you quoted cannot
> be thrown away,
>
> Counterexample anyone?
How about the limit as x --> infinity of
x / [x - sin(x)],
or equivalently (let x = 1/u),
the limit as u --> 0+ of
1 / [1 - u*sin(1/u)] ?
It's easy to show the limit is 1, but it seems
you'll run into trouble trying to show this using
l'Hopital's rule.
Dave L. Renfro
ba...@barrs.org
for a condition. Even if the condition is redundant or unnecessary, it
is not wrong--not a meathematical error--to put it there. For example,
in defining an analytic function, one commonly supposes that the real
and imaginary parts are of class C^1, when considerably less will do.
However, it would be wrong pedagogically to start a course in complex
variables by getting into all that. Or the definition of a Lie group
only requires a topological group structure (continuous multiplication
and inverse) on a manifold and analyticity follows (that's Hilbert's
fifth problem). But that is a deep and difficult problem and when I
took a course in topological groups, class C^2 was assumed.
Now in the case at hand, it is not inconceivable that if you stick to
the set on which g(x) is non-zero and keep the remaining hypotheses,
that the L'Hospital's rule will work along that set. I have never
tried that, but the point is that that would be an extension of the
usual statement.
Robert Israel
What is this a counterexample to? The subject under
discussion was whether you need to assume g(x) <> 0
in some neighbourhood of a for l'Hopital's Rule to
be valid as x -> a, not whether you need g'(x) <> 0.
Dave L. Renfro
> What is this a counterexample to? The subject under
> discussion was whether you need to assume g(x) <> 0
> in some neighbourhood of a for l'Hopital's Rule to
> be valid as x -> a, not whether you need g'(x) <> 0.
I was in a hurry (I was at work, had a little free
time, and then something came up), but the function
I gave is zero in every neighborhood of oo (or 0,
for the re-variabilized example). However, I see
that it isn't a counterexample in the sense you're
looking for since l'Hopital doesn't even work for
the example I gave. I think you want an example where
g(x) = 0 for a sequence approaching x=a, f'(x)/g'(x)
has a limit as x --> a, but f(x)/g(x) doesn't have
a limit as x --> a (where the limit is taken along
the set where g(x) isn't zero), right? If I get a
chance (meaning, if I remember to), I'll dig around
some stuff I have on l'Hopital's rule and see if I
find anything relevant.
Incidentally, there's a more general version of
l'Hopital's rule involving lim-inf and lim-sup
that was first (I think) published by William
H. Young in 1910 (and rediscovered several times
by others). I don't think it's directly relevant to
the discussion here, but it might be of interest.
L'Hopital's rule says that, under appropriate hypotheses,
f'/g' --> L implies f/g --> L. However, the converse
doesn't have to hold. In fact, it's possible for f/g
to have a limit, but not f'/g', and [x - (sin x)] / x
is an example. However, it is true that
lim-inf(f'/g') \leq lim-inf(f/g) \leq lim-sup(f/g) \leq lim-sup(f'/g').
In fact, Young proved that every subsequential limit point
of f/g is a subsequential limit point of f'/g'.
Young's paper is cited in a recent post of mine, at
http://groups.google.com/group/alt.math.undergrad/msg/eb8efd19eebab8f0
Dave L. Renfro
The World Wide Wade
<1136935664.6...@g47g2000cwa.googlegroups.com>,
"Dave L. Renfro" <renf...@cmich.edu> wrote:
> I was in a hurry (I was at work, had a little free
> time, and then something came up), but the function
> I gave is zero in every neighborhood of oo
??? The function you gave is x / [x - sin(x)].
Luke Wu
Makes sense. Thank you and others.