L'Hopital's Rule in 2-D ?
Terry Anderson
variables?
Chris Campbell
cc6...@conrad.appstate.edu
Calculus III student, posted with instructor's knowledge (on
instructor's account)
--
Terry Anderson t...@math.appstate.edu
Math Sciences Dept. Appalachian State University
Boone, NC 28608 USA (704) 262 - 2357
t...@flyer.ncsc.org t...@cardinal.ncsc.org
Kazimir Kylheku
Terry Anderson <t...@cs.cs.appstate.edu> wrote:
>Is there a version of L'Hopital's Rule for two or more independent
>variables?
I have given this some thought, and it appears that a fruitful
direction to take is to switch everything to polar coordinates.
For instance, the limit of 2xy/(x^2 + y^2) as (x,y) -> (0,0) doesn't
exist, while the limit of 2x^2y/(x^2 + y^2) does exist, and is zero.
The former expressed in polar form about (0,0) is:
2(rcos@)(rsin@)/r^2 = 2cos@sin@ = sin(2@)
This limit obviously depends on the value of the angle @.
Now the latter formula expressed in polar coordinates is:
2(r^2cos^2@)(rsin@)/r^2 = rcos^2@sin@
This can be directly evaluated as zero at r = 0 for any @, and thus
the limit is zero.
Now let's try xy/(x^3 + y):
-> r^2cos@sin@/(r^3cos^3@ + rsin@)
-> rcos@sin@/(r^2cos^3@ + sin@) /* cancel some r's */
The directional limit is zero from only the directions that have a
non-zero sin@, otherwise it's undefined. Thust the general limit
doesn't exist.
I'm sure there must be an example where L'Hopital's rule can be
applied as partial derivatives in r.
How about this beast:
f(x,y) = xy / sin(x + y)
off to polar with it!
--> r^2cos@sin@ / sin[r(cos@ + sin@)]
Now what? At (0,0), both numerator and denominator are zero. Treating
the angle as constant, we differentiate with respect to r, L'Hopital style:
--> 2rcos@sin@ / { (cos@ + sin@) cos[r(cos @ + sin@)] }
The limit would appear to be 0 except that along the line x = -y, the
factor (cos@ + sin@) is zero, and the limit is undefined.
It appears that you CAN use L'Hopital, as long as you separate the
distance and direction by using polar coordinates.
I'd like to see an example where L'Hopital actually gives an answer,
rather than merely reducing the function to an easy-to-analyze form.
Let's try:
lim|(x,y)->(0,0) (x^2 + y^2)/sin(x^2 + y^2)
This of course, is just
r^2/sin(r^2)
which can be solved by L'Hopital
-> 2r / cos(r^2) -> 0/1 -> 0
It's now totally obvious to me how L'Hopital's rule can be applied in
the solution multi-variable limits.
Dik T. Winter
> In article <39qu25$1...@lester.appstate.edu>,
> Terry Anderson <t...@cs.cs.appstate.edu> wrote:
> >Is there a version of L'Hopital's Rule for two or more independent
> >variables?
> I'm sure there must be an example where L'Hopital's rule can be
> applied as partial derivatives in r.
Yes, *if* the limit exists and is independent of the path along which
you calculate the limit, you can take any path, apply any change of
coordinates and find the limit. Some may need L'Hopital's Rule, some
not. But whatever you do, L'Hopital's Rule is a rule in one variable,
you do not supply a generalization in more variables.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924098
home: bovenover 215, 1025 jn amsterdam, nederland; e-mail: d...@cwi.nl
Kazimir Kylheku
Kazimir Kylheku <c2a...@ugrad.cs.ubc.ca> goofed up:
>This of course, is just
>
> r^2/sin(r^2)
>
>which can be solved by L'Hopital
>
>-> 2r / cos(r^2) -> 0/1 -> 0
As was kindly pointed out to me in e-mail, I screwed up the old
chain-rule there. The bottom derivative should of course be 2r
cos(r^2), and the limit should of be 1.
Cheers..
Kaz
Ronald Bruck
>In article <39sbk2...@bowen.ugrad.cs.ubc.ca> c2a...@ugrad.cs.ubc.ca (Kazimir Kylheku) writes:
> > In article <39qu25$1...@lester.appstate.edu>,
> > Terry Anderson <t...@cs.cs.appstate.edu> wrote:
> > >Is there a version of L'Hopital's Rule for two or more independent
> > >variables?
>...
> > I'm sure there must be an example where L'Hopital's rule can be
> > applied as partial derivatives in r.
>
>Yes, *if* the limit exists and is independent of the path along which
>you calculate the limit, you can take any path, apply any change of
>coordinates and find the limit. Some may need L'Hopital's Rule, some
>not. But whatever you do, L'Hopital's Rule is a rule in one variable,
>you do not supply a generalization in more variables.
Perfectly true, but L'H. in several variables doesn't look very useful to
me anyway, and I'll explain why.
Suppose we have functions f, g mapping an open subset U of a Banach space
to R, which are Gateaux-differentiable at x \in U, and such that f(x) =
g(x) = 0. We want to know whether $\lim_{y \to x} f(y)/g(y)$ exists, and
what its value is.
Obviously a NECESSARY condition is that for any h in the Banach space,
$\lim_{t \to 0} f(x+th)/g(x+th)$ exists and takes a common value $c$
independent of $h$. (Hmmm, my TeX is running wild. Pretty soon I'll be
speaking it aloud.)
Applying L'Hopital's rule for functions of a single variable, this will
require that <f'(x),h> / <g'(x),h> = c for all h not in the kernel of
g'(x), i.e. that <f'(x),h> = c <g'(x),h> for all h not in the kernel of
g'(x). If g'(x) is not the zero functional, this obviously implies that
f'(x) = c g'(x). And if g'(x) IS the zero functional, then pretty clearly
f'(x) will have to be the zero functional too.
So for this form of L'Hopital's rule to be viable, you have to be doing it
at a point where f'(x) and g'(x) are linearly dependent. That's a pretty
strong statement for functions of several variables. Haven't thought
about whether this condition is SUFFICIENT too--almost certainly not with
just Gateaux-differentiability, which is only about directional
derivatives, but I suspect it is if f and g are Frechet-differentiable at
x with f(x) = g(x) = 0.
--Ron Bruck
Pertsel Vladimir
In article <39qu25$1...@lester.appstate.edu>, t...@cs.cs.appstate.edu (Terry Anderson) writes:
|> Is there a version of L'Hopital's Rule for two or more independent
|> variables?
Why do You need a special one (L'Hopital's Rule is applicable to
the partial derivatives)? And besides that, how can it look like?
If the ratio of the partial derivatives it the same --- than
there exists a limit of a fraction of two infinitely small
functions. If not --- then the limit doesn't exist. (Under the
assumption, that all the partial derivatives are continuous in a
n-dimensional neighbourhood of the point.)
--
/\ /\ Vladimir A. Pertsel
((ovo)) Tel:(972) 03-5600253 (res.) 08-434303 (bus.)
():::() URL ftp://ftp.wisdom.weizmann.ac.il/pub/voldemar/pertsel.html
PVA E-mail: vold...@wisdom.weizmann.ac.il
la...@inland.com
> In article <39sbk2...@bowen.ugrad.cs.ubc.ca> c2a...@ugrad.cs.ubc.ca (Kazimir Kylheku) writes:
> > In article <39qu25$1...@lester.appstate.edu>,
> > Terry Anderson <t...@cs.cs.appstate.edu> wrote:
> > >Is there a version of L'Hopital's Rule for two or more independent
> > >variables?
> ...
> > I'm sure there must be an example where L'Hopital's rule can be
> > applied as partial derivatives in r.
>
> Yes, *if* the limit exists and is independent of the path along which
> you calculate the limit, you can take any path, apply any change of
> coordinates and find the limit. Some may need L'Hopital's Rule, some
> not. But whatever you do, L'Hopital's Rule is a rule in one variable,
> you do not supply a generalization in more variables.
There is a way to generalize the rule to more than one variable, but in
practice it can only be applied in one direction at a time. To wit, suppose
that as x -> x_0 (x is considered a coordinate vector here), we have f -> 0
and g -> 0. If lim {x -> x_0} (f/g) exists and is equal to L and v is any
nonzero vector in the coordinate space, then:
L = lim {x -> x_0} [v.grad f/(v.grad g)]
Dotting the gradients with v amounts to approaching from one direction for the
given v, hence the "multivariate" L'Hopital's rule can be applied in only one
direction at a time. Note that the above formula provides a condition for the
existence of the limit: the right side must exist for all nonzero v and be
independent of v.
--OL
la...@inland.com