L'Hopital for R^n
William Elliot
Is there any extension of/version of L'Hopital's rule for R^2; for
R^n in general?. In particular, I was trying to deal with the problem
of finding the limit at (0,0) of
|xy|
____
(x^2+y^2)^q
q in [0,1]
The value near (0,0) seemed to lead to limits of the form 0/0, so I
tried to play around with the gradients above and below. It seemed to
work, but I am not sure if it is just coincidence.I was basically
trying to bound the norm of the partial above and below separately.
Does this make sense?. Is there some other generalization, maybe
somewhat modified? I would appreciate comments. Thanks.
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David C Ullrich
>Posted by: Farfan at Ask An Analyst
>Is there any extension of/version of L'Hopital's rule for R^2; for
>R^n in general?.
I tend to doubt it.
>In particular, I was trying to deal with the problem
>of finding the limit at (0,0) of
>|xy|
>____
>(x^2+y^2)^q
>q in [0,1]
>The value near (0,0) seemed to lead to limits of the form 0/0, so I
>tried to play around with the gradients above and below. It seemed to
>work, but I am not sure if it is just coincidence.
If you say exactly what "it" is people would have a better chance
of saying whether "it" is a correct result or not
>I was basically
>trying to bound the norm of the partial above and below separately.
>Does this make sense?. Is there some other generalization, maybe
>somewhat modified? I would appreciate comments. Thanks.
>
>----
>
>
>
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David C. Ullrich
Martin Cohen
> Posted by: Farfan at Ask An Analyst
> Is there any extension of/version of L'Hopital's rule for R^2; for
> R^n in general?. In particular, I was trying to deal with the problem
> of finding the limit at (0,0) of
> |xy|
> ____
> (x^2+y^2)^q
> q in [0,1]
> The value near (0,0) seemed to lead to limits of the form 0/0, so I
> tried to play around with the gradients above and below. It seemed to
> work, but I am not sure if it is just coincidence.I was basically
> trying to bound the norm of the partial above and below separately.
> Does this make sense?. Is there some other generalization, maybe
> somewhat modified? I would appreciate comments. Thanks.
y = k x^v) for constant k. In this case, I get
kx^2/(x^(2q)*(1+k^2)^q) = x^(2-2q)*k/(1+k^2)^q.
If 0 < q < 1, this -> 0 as x -> 0 (if x > 0).
As for the partials, I feel lasy. Maybe convert to polar form
(x = r cos t, y = r sin t, x^2 + y^2 = r^2).
Martin Cohen
Lynn Kurtz
wrote:
Of course, showing that the limit is 0 along all those paths doesn't
prove the limit as (x,y)->(0,0) is zero for 0<q<1, even though it is.
But by expanding
(|x|-|y|)^2 >= 0
you can show |x||y| <= (1/2)*(x^2+y^2), from which it does follow.
When q=1 your technique can show the limit fails to exist.
--Lynn
James B. Sibley
> Posted by: Farfan at Ask An Analyst
> Is there any extension of/version of L'Hopital's rule for R^2; for
> R^n in general?. In particular, I was trying to deal with the problem
> of finding the limit at (0,0) of
> |xy|
> ____
> (x^2+y^2)^q
> q in [0,1]
> The value near (0,0) seemed to lead to limits of the form 0/0, so I
> tried to play around with the gradients above and below. It seemed to
> work, but I am not sure if it is just coincidence.I was basically
> trying to bound the norm of the partial above and below separately.
> Does this make sense?. Is there some other generalization, maybe
> somewhat modified? I would appreciate comments. Thanks.
I have a strong feeling this deals with calculus of multivariables. I
tried my hand at a two variable limit which dealt with a problem my
classes was having (poor souls). We were looking for vertical tangents
of a derivative in implicit form. One of the values was x = 1, however
the derivative gave 0/0 (which was with respect to x and y). I could
not figure out the double variable limit because I did not know at
what rate y changed with x at any point x (I suppose that is the
correct logic). With my approach, dy/dx->3 as x,y->(1,2). The actual
derivative was sqrt(3) which I obtained after manipulating the
original equation. So basically, I don't think one can evaluate this
limit unless more is known.
I could be wrong because I am not a student of multivariable calculus.
If I am wrong, someone please help me :)
James Sibley
Athanasios
I think it is obvious that:
|xy|
____ is always less than or equal to 1/2*(x^2+y^2)^(1-q) which
(x^2+y^2)^q
tends to zero as x->0 and y->0.
Therefore, the required limit is 0.
Thanos
Stephen J. Herschkorn
>
>Is there any extension of/version of L'Hopital's rule for R^2; for
>R^n in general?. In particular, I was trying to deal with the problem
>of finding the limit at (0,0) of
>|xy|
>(x^2 +y^2)^q
>q in [0,1]
>The value near (0,0) seemed to lead to limits of the form 0/0, so I
>tried to play around with the gradients above and below. It seemed to
>work, but I am not sure if it is just coincidence.I was basically
>trying to bound the norm of the partial above and below separately.
>Does this make sense?. Is there some other generalization, maybe
>somewhat modified? I would appreciate comments. Thanks.
>
Letting f be a real-valued function on a neighborhood of 0 in R^n,
if lim f(x) exists, then so does lim f(te) and the two limits are
equal, where x is a vector variable, e is a constant vector, t
is a scalar variable, and the limits are as x and t approach 0.
[In the sequel, all limits are taken as the argument approaches 0.]
So, if lim f(x)/g(x) exists and lim f(x) = lim g(x) = 0, then
L'Hospital tells us that the limit of the quotient is equal to the limit
of quotient of partials w.r.t the same variable if the latter limit
exists. That is, the multidimensional limit, if it exists, reduces to a
one-dimensional along one of the coordinate axes. Similarly, one could
use directional derivatives.
So the problem becomes on of determining when the limit exists. In the
submitted example, the limit does *not* exist when q = 1. For,
denoting the quotient in this case by f(x,y), lim f(x,0) = lim f(0,x)
= 0, whereas lim f(x,x) = 1/2. My guess is that the limit does exist
for q<1, but this fact requires proof. I suspect that there is no
L'Hospital's rule for the establishment of the existence of
multidimesional limits, for none of Bartle, Buck, Rudin, or Spivak
mention such a theorem..
By the way, the agreement of one-dimensional limits along fixed
directions is a necessary condition for the existence of the
multidimensional limit, but this condition is not sufficient. A
necessary and sufficient condition is that the limits of f(g(t)) agree
for all functions g: (0,1) -> R^n (domain is the open unit interval)
such that lim g(t) = 0.
--
Stephen J. Herschkorn hers...@rutcor.rutgers.edu
Stephen J. Herschkorn
>
>
>I think it is obvious that:
> |xy|
> ____ is always less than or equal to 1/2*(x^2+y^2)^(1-q) which
> (x^2+y^2)^q
>
>tends to zero as x->0 and y->0.
>Therefore, the required limit is 0.
>
As long as q <1. (The original post had 0 <= q <= 1.) Good
observation, though, respectfully, the characterization "obvious" tends
to insult the reader here.
Also, this observation does not address the existence of a
multidimensional L'Hospital's rule, which was the point of the original
post.
Best regards,
Dave L. Renfro
[sci.math Jan 4 2003 4:30:29:000PM]
http://mathforum.org/discuss/sci.math/m/470587/470645
wrote (in part, replying to Martin Cohen):
>> For these type of limits, try setting y = kx (or, possibly,
>> y = k x^v) for constant k.
>
> Of course, showing that the limit is 0 along all those
> paths doesn't prove the limit as (x,y)->(0,0) is zero
> for 0<q<1, even though it is.
I recently gave a simple example of this in alt.math.undergrad
that might be of interest to those reading the present thread.
A copy of that post (slightly edited) is given below.
Dave L. Renfro
--------------------------------------------------------------------
http://mathforum.org/epigone/alt.math.undergrad/crantendbong/ze0epc1fnf8i@legacy
Subject: Re: Evaluation of multivariable limit
Author: Dave L. Renfro <renf...@cmich.edu>
Date: 1 Jan 03 21:10:41 -0500 (EST)
G.E. Ivey <georg...@gallaudet.edu>
[alt.math.undergrad 1 Jan 03 09:02:44 -0500 (EST)]
http://mathforum.org/epigone/alt.math.undergrad/crantendbong
wrote (in part):
> In order to show that a limit one must show that one gets the
> same result approaching (a,b) on any curve (approaching on a
> line or curve reduces to the one-variable problem). Even
> looking at all straight lines is not enough. Most calculus
> books give example of functions where one gets the same result
> approaching along any STRAIGHT LINE, but a different result
> approaching along a parabola.
Here's a simple function that's even worse -->
Let f(x,y) = y^x for y > 0. Then f has the same limit for any
polynomial approach to (0,0) from the upper half plane (if you
try to verify this you'll see that it will suffice in this case
to consider curves of the form y = ax^n), but the limit will vary
for approach paths of the form y = exp(-b/x) for b > 0.
Note that all the of y = exp(-b/x) curves are approaching
(0,0) faster than any of the y = ax^n curves in the sense
that lim(x --> 0+) of [exp(-b/x)] / (x^n) is equal to 0
for all positive integers n and real numbers b > 0. Moreover,
as b varies, these exponentially behaved curves themselves
have varying relative rates of approach that are greater
than the varying relative rates of approach between any
two powers of x. [If b1 < b2 and you divide exp(-b1/x) by
exp(-b2/x), then you'll get exp[(b2-b1)/x], whose behavior
near x=0 dwarfs the behavior you get when you divide x^m
by x^n for any m < n.]
Because y^x tends to be a lot more sensitive to changes in x
than changes in y, it makes sense to look at approach paths
that are nearly horizontal. However, f is apparently so
uniformly behaving near the origin that polynomial paths
are not sufficiently different among themselves to show
different limiting behavior in f at the origin.
Dave L. Renfro
--------------------------------------------------------------------
David C Ullrich
wrote:
Well, saying that it has to do with multi-variable calculus is not
wrong, it's just obvious, like saying you have a strong feeling
that Fermat's last theorem has to do with number theory.
Your suggestion that the limit cannot be evaluated is certainly
wrong:
Note that |x| <= (x^2 + y^2)^(1/2) and similarly for |y|,
so |xy| <= (x^2 + y^2). Hence
(1) 0 <= |xy|/(x^2+y^2)^q <= (x^2 + y^2)^(1-q).
If q < 1 then 1 - q > 0, so (1) shows that the given function
has limit 0 as (x,y) -> (0,0).
On the other hand, if q = 1 then the limit does not exist.
Let's say f(x,y) = |xy|/(x^2+y^2)^q = |xy|/(x^2+y^2).
Then f(x,0) = 0 for all x <> 0, so f has limit 0 as you
approach the origin along the y-axis. On the other
hand f(t,t) = 1/2, so f has limit 1/2 as you approach
the origin along the line x = y. Since f has two different
limits along these lines approaching the origin the
limit as (x,y) -> 0 of f(x,y) does not exist.
And there you are - I've evaluated the limit for
every q in [0,1].
>If I am wrong, someone please help me :)
>
>James Sibley
David C. Ullrich
James B. Sibley
> Well, saying that it has to do with multi-variable calculus is not
> wrong, it's just obvious, like saying you have a strong feeling
> that Fermat's last theorem has to do with number theory.
> Your suggestion that the limit cannot be evaluated is certainly
> wrong:
>
> Note that |x| <= (x^2 + y^2)^(1/2) and similarly for |y|,
> so |xy| <= (x^2 + y^2). Hence
>
> (1) 0 <= |xy|/(x^2+y^2)^q <= (x^2 + y^2)^(1-q).
>
> If q < 1 then 1 - q > 0, so (1) shows that the given function
> has limit 0 as (x,y) -> (0,0).
>
> On the other hand, if q = 1 then the limit does not exist.
> Let's say f(x,y) = |xy|/(x^2+y^2)^q = |xy|/(x^2+y^2).
> Then f(x,0) = 0 for all x <> 0, so f has limit 0 as you
> approach the origin along the y-axis. On the other
> hand f(t,t) = 1/2, so f has limit 1/2 as you approach
> the origin along the line x = y. Since f has two different
> limits along these lines approaching the origin the
> limit as (x,y) -> 0 of f(x,y) does not exist.
>
> And there you are - I've evaluated the limit for
> every q in [0,1].
>
> >If I am wrong, someone please help me :)
> >
> >James Sibley
>
>
> David C. Ullrich
Thank you Mr. Ullrich for your answer. It makes sense. However, I have
a question.
> >I could
> >not figure out the double variable limit because I did not know at
> >what rate y changed with x at any point x (I suppose that is the
> >correct logic). With my approach, dy/dx->3 as x,y->(1,2). The actual
> >derivative was sqrt(3) which I obtained after manipulating the
> >original equation.
Is my logic correct in this case?
I was going through my "organized" stack of papers and found the
problem I was speaking of. To deprive you, I will only give you the
derivative (hehe.. please no integration). which is as follows:
dy/dx = 3*(x-1)^2 / [2 * (y - 3) ]
The purpose was to find the vertical asymptotes. The problem point is
(1, 3). dy/dx|{x,y}=(1,3) = 0/0. Question, is it vertical or not? As I
found out later, the actual value approaches +-sqrt(3) (depending on
how you approach it).
James Sibley
The World Wide Wade
renf...@cmich.edu (Dave L. Renfro) wrote:
> > In order to show that a limit one must show that one gets the
> > same result approaching (a,b) on any curve (approaching on a
> > line or curve reduces to the one-variable problem). Even
> > looking at all straight lines is not enough. Most calculus
> > books give example of functions where one gets the same result
> > approaching along any STRAIGHT LINE, but a different result
> > approaching along a parabola.
>
> Here's a simple function that's even worse -->
>
> Let f(x,y) = y^x for y > 0. Then f has the same limit for any
> polynomial approach to (0,0) from the upper half plane (if you
> try to verify this you'll see that it will suffice in this case
> to consider curves of the form y = ax^n), but the limit will vary
> for approach paths of the form y = exp(-b/x) for b > 0.
Worse in some ways, but you're cheating just a little: your f is not
defined in any punctured disc centered at (0,0). We could try this: define
f(x,y) = 1 if y = exp(-1/x^2), 0 otherwise. Then f has limit 0 on any
polynomial path through the origin, but by design has limit 1 on the curve
y = exp(-1/x^2). Well that's cheating too, because this f is highly
discontinuous. It would be easy enough to smooth this f out to be C^oo
however. Your f is real-analytic on the upper half plane, but there is no
way to extend it smoothly to a punctured neighborhood of (0,0).
Suppose we try this: For (x,y) in R^2 \ {(0,0)}, set
f(x,y) = [y - exp(-1/(x^2+y^2)]^2 / [y^2 + exp(-1/(x^2+y^2)].
Then f is real-analytic on R^2 \ {(0,0)}. Now suppose p(t) is a nonzero
path in R^2 that is polynomial in both coordinates, and p(0) = (0,0). Then
f(p(t)) -> 1 as t -> 0, as a little work shows. Yet f equals 0 on the level
set y = exp(-1/(x^2+y^2), which has (0,0) as a limit point. Therefore f
fails to have a limit at (0,0).