L'Hopital and Multivariable Calculus
Mike Stump
Calc) class today.. As we probably all know, L'Hopital's rule states that
when we're taking the limit of a quotient which yields an indeterminate form
when we substitute the value in, such as
lim(as x->0) of (x/x^2) *which yields 0/0*,
we can take the derivative of the numerator and the derivative of the
denominator, and take the limit then. *To use the above ex.,
lim(as x->0) of (x/x^2) = lim(as x->0) of (1/2x) = 1/(2*0) = 1/0 = infinity
Well, we were discussing functions of three independent variables x, y and z
today, and the question arose- If we were to take the
lim(as (x,y,z)->(a,b,c)) of (f(x,y,z))
and when we substitute to get f(a,b,c) we get an indeterminate value (such
as 0/0 or infinity/infinity),
is there another form of L'Hopital's Theorem which covers multivariable
cases? I'm sure that if it exists it'd be quite laborious and involve
partial deriv's up the gazoo, but it could come in handy.
If any of you can help, please email me at mstu...@student.villanova.edu,
and myself and my professor will be extremely grateful. Thanks a bunch!
---Michael Stump
2nd Yr. Astrophysics Major, Villanova U. (Villanova, PA)
William C Waterhouse
Mike Stump" <mstu...@student.vill.edu> writes:
>...
> today, and the question arose- If we were to take the
>
> lim(as (x,y,z)->(a,b,c)) of (f(x,y,z))
>
> and when we substitute to get f(a,b,c) we get an indeterminate value (such
> as 0/0 or infinity/infinity),
>
> is there another form of L'Hopital's Theorem which covers multivariable
> cases?
The fact is that such expressions almost never have limits, first
because they aren't defined in the full neighborhood of the point
and second because they won't have a limit even where they are
defined.
Both problems are visible in the simplest case in two variables,
[ax + by]/[cx + dy]
near the origin. First, cx+dy = 0 gives a whole line through the
origin where the ratio isn't defined. Secondly, dividing numerator
and denominator by x, we see that this ratio is equal to
[a + b(y/x)]/[c + d(y/x)].
Thus if you approach the origin along lines of different slopes,
you get different limiting values.
William C. Waterhouse
Penn State
bo...@rsa.com
Yup.
Allow me to state this somewhat differently.
In multiple dimensions the value of a function at a limit point usually
depends on the PATH you take in approaching the limit point. Approach
along one path and you get one answer. Approach along another and
you get another answer. In the univariate case (1 dimension) this problem
does not arise because there is only one path to take to approach the limit.
The concept of "analytic function" in the complex plane arises precisely
because of this difficulty.
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Jeff Bertolet
limits you can make substitutions to simplify.
lim (x,y)--> (0,0) (2x^2*y)/(x^4 + y^2)
make the subst. y=mx, which means you approach along some straight line.
lim x--> 0 2mx^3/(x^4 + m^2x^2)= lim x--> 0 2mx/(x^2+m^2)=0
BUT,
make subs. y=x^2, meaning you approach along parabola.
lim x--> 0 2x^4/(x^4+x^4)=1
DWCantrell
>In multiple dimensions the value of a function at a limit point usually
>depends on the PATH you take in approaching the limit point.
Replace "usually" with "generally" and I would then agree.
>Approach
>along one path and you get one answer. Approach along another and
>you get another answer. In the univariate case (1 dimension) this problem
>does not arise because there is only one path to take to approach the limit.
False. In one dimension, there are normally two paths; they can be called
"approach from the left" and "approach from the right". Of course, limits from
left and right need not agree.
Cheers,
David Cantrell