L'Hopital's Rule...
Joni Concilio
up, I've encountered a question which we never dealt with in high school calc
(for obvious reasons).
When using L'Hopital's Rule to evaluate a limit where the result is 0/0
(undefined), when specifically will taking the derivative of the top and bottom
as per L'Hopital not work? I remember our high school book saying that there
were cases where using L'Hopital's Rule would cause a "loop", where even using
the rule repeatedly would not work, but am unable to come up with a case in
which this would happen. Maybe I'm just lost... but could anyone help?
.-*~Joni Concilio~*-.
"If we knew all the answers, then we wouldn't be scientists, but gods."
deanl...@worldnet.att.net
Joni Concilio wrote in message
<19990819141743...@ng-fe1.aol.com>...
No 0/0 forms like this come to mind immediately, but there is a pretty
simple instance of an (infinity)/(Infinity) form like this where L'Hopital
alone doesn't quite cut it.
Consider limit as x -> infinity of x/Sqrt[x^2+1].
One application of LH shows this should be equal to limit as x-> infinity of
Sqrt[x^2+1]/x. Presumably you can guess what the second application of LH
yields.
Of course, if you know some other things, are a bit clever and first show
that this limit exists, the first application of LH can be used to evaluate
this limit.
Dean
Adam Bryant
"L'Hopital loop". Consider limit as x->0 of sin(x)/cos(x+pi). Apply
L'Hopital's Rule a few times and the loop is obvious. This example is
degenerate because sin(x) is identically cos(x+pi), so you know the
limit is 1 without L'Hopital's Rule. I'm a bit puzzled why you didn't
study this in high school Cal 2 since we covered it in high school Cal
1.
Adam Bryant
Adam Bryant
rather than sin(x)/cos(x+pi). Also, it should read sin(x) is
identically cos(x+pi/2). Instead of proofreading, I guess I hoped spell
check would catch my mistyped functions :).
Adam Bryant
Dave L. Renfro
<http://forum.swarthmore.edu/epigone/alt.math.undergrad/snangterdmul/19990819141743...@ng-fe1.aol.com>
writes:
>Hi! I'm a college freshman getting ready to start Calc 3,
>and while brushing up, I've encountered a question which
>we never dealt with in high school calc (for obvious reasons).
"for obvious reasons"?? It certainly came up in the
high school classes I've taught the past 3 years ...
>When using L'Hopital's Rule to evaluate a limit where
>the result is 0/0 (undefined), when specifically will taking
>the derivative of the top and bottom as per L'Hopital not work?
>I remember our high school book saying that there were cases
>where using L'Hopital's Rule would cause a "loop", where
>even using the rule repeatedly would not work, but am unable
>to come up with a case in which this would happen.
>Maybe I'm just lost... but could anyone help?
>
>.-*~Joni Concilio~*-.
>"If we knew all the answers, then we wouldn't be scientists,
>but gods."
Try limit(x --> 0+) of [exp(-x^(-1)]/x.
More generally,
limit(x --> 0+) of [exp(-a*x^(-1)] / [x^b], where
a and b are any positive real constants.
A direct application of L'Hopital's rule using the
given quotient form gets you into an infinite loop
of quotients that get further and further from where
you'd like to go.
Here are a couple of ways I can think of that
will work.
METHOD 1: Reverse the quotient, and apply L'Hopital to
limit(x --> 0+) of [x^(-1)] / [exp(x^(-1)].
METHOD 2: Make the variable change u <---> x^(-1).
Since x --> 0+ if and only if u --> infinity,
the limit transforms under the variable change to
limit(u --> infinity) of u / [exp(u)], whose
evaluation is now immediate.
By the way, I hope you know these cautions when
working with L'Hopital's rule:
CAUTION 1: Only apply L'Hopital's rule to 0/0 or
infinity/infinity forms. For example, if you
"do L'Hopital" to
limit(x --> 0) of (x+1)/(x+2),
you'll get the incorrect value 1.
CAUTION 2: L'Hopital's rule actually has the form of a
(one-way) conditional: If a "L'Hopital
differentiated limit exists", then the
original limit exists and its value is
the same as the "L'Hopital differentiated
limit". [I'm assuming the necessary
differentiability conditions for L'Hopital
(which are probably a bit more than just the
existence of the appropriate derivatives at
the point) are satisfied, of course.]
There are examples in which the limit
actually exists, but the "L'Hopital
differentiated limit" doesn't exist. I don't
have an example now (I'm still unpacking
from moving to a different state a few days
ago, so my stuff is in disarray), but my
recollection is that you need to incorporate
a function that changes sign infinitely often
(or maybe its derivative changes sign
infinitely often) in a neighborhood of the
point [e.g. sin(1/x) or (x^2)*sin(1/x)
behavior in a nbd. of x=0].
Joni Concilio
you who are thinking that my high school has a "degenerate case" of a math
department, I only had one semester of high school calculus... and in the words
of my calculus teacher "Unless you're going to be a math major.... ((followed
by whatever it was we didn't need to know)) and if you ARE (once I told her
that in fact I was going to be a math major) you will learn this later!"
I made college Calc 3 with a lot of studying and some luck on the Calculus
Advanced Placement exam. I've been reviewing a lot of things that maybe some
would have gone over in calc 2. That's where L'Hopital and such things have
come from.
Thanks again for all your help... I'll keep checking posts now that I know
there are other "weird" people who don't mind talking about math! (I'm only
kidding. I'm weird, but it doesn't have anything to do with math.)
See ya!
.-*~Joni Concilio~*-.