Can we use L'hopital rule here?
lihuijun
0/s as s-->0 (for positive 0)
0/(s^2) as s-->0 (for positive 0)
If so, then the ans is 0. Please advise
Regards
anon
I don't know mathematics with positive zero. You probably mean
positive s. Then, you need the limit of the following function:
f(s) = 0/s when s tends to zero from above. However, no matter how small
s>0 is, the above expression equals zero so the limit is zero(use
the definition of limit).
Thomas Nordhaus
Yes, you can. Let f(s) = 0 for all s. What is f(0)? What is f'(0)?
f''(0)?
Thomas
>
>Regards
The World Wide Wade
lihu...@hotmail.com (lihuijun) wrote:
> Can we use L/hopital rule here?
>
> 0/s as s-->0 (for positive 0)
>
> 0/(s^2) as s-->0 (for positive 0)
Yes, but it would be idiotic.
Ronald Bruck
<lihu...@hotmail.com> wrote:
When did YOU quit beating your wife?
Your question makes sense (if it is rather foolish) until you
append "(for positive 0)". I don't know what you mean, and I
can't even guess what you might mean.
Perhaps you mean "f(s)/s, where f(s) --> 0+" (making the beginner's
mistake of substituting one of the limits in the expression f(s)/g(s),
leaving the other unsubstituted). Problem with this interpretation:
why would it make any difference that f(s) approaches 0 from the
POSITIVE side?
Perhaps you mean "0/s, where s --> 0+". Then the answer is yes. But,
as I said, it's foolish, because 0/s is identically 0 for s > 0, so its
limit is also 0.
Did your professor count you WRONG for using L'Hospital's rule? Then he
was, IMHO, absolutely correct.
--Ron Bruck
Robin Chapman
L'hopital's rule is rarely useful, but this must be
the most pointless application of it ever devised!
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9"
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
Bill Dubuque
>lihuijun <lihu...@hotmail.com> wrote: (edited)
>>
>> Can we use L/hopital rule here?
>>
>>
>> 0/x^2 as x-->0
>>
>> If so, then the ans is 0. Please advise
>
> L'hopital's rule is rarely useful, but this must be
> the most pointless application of it ever devised!
Before rushing to judgment, consider the following
2 2
sin (x) + cos (x) - 1
lim ---------------------
x->0 x^2
The expression equals 0/x^2, so has the above form.
Applying L'Hopital's rule reduces it to the trivial
limit of 0/(2x) and thus yields the correct result
_without_ requiring the specialized trig knowledge
that the initial numerator is identically zero.
Now imagine a similar limit only with a much more
hairy numerator that is identically zero but is not
immediately recognizably so. As above, L'Hospital's
rule might similarly simply reveal the correct result.
The same remark apply to machines as well as humans.
Computer algebra systems may encounter such limits.
Further, suppose that the rule didn't apply in the
case the numerator is identically zero. Then before
applying the rule one would be required to verify
also that the numerator was not identically zero.
But the zero equivalence problem is undecidable
in general (even so for most any nontrivial class
of functions one encounters in calculus, see [1]).
So it's a good thing the rule applies in this case.
--Bill Dubuque
[1] Daniel Richardson: Some Undecidable Problems Involving Elementary
Functions of a Real Variable. J. Symb. Log. 33(4): 514-520 (1968).
The World Wide Wade
Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
> Before rushing to judgment, consider the following
>
> 2 2
> sin (x) + cos (x) - 1
> lim ---------------------
> x->0 x^2
>
>
> The expression equals 0/x^2, so has the above form.
> Applying L'Hopital's rule reduces it to the trivial
> limit of 0/(2x) and thus yields the correct result
> _without_ requiring the specialized trig knowledge
> that the initial numerator is identically zero.
That's dubious for two reasons. First, to apply LHR, you must verify the
numerator -> 0; how will you do that without knowing sin^2 + cos^2 = 1?
Second, we're to believe someone has forgotten the most basic of trig
identities, sort of like forgetting your own name, and yet knows how to
apply the chain rule and remember the derivatives of sin and cos correctly?
Well, having taught enough calculus by now, I suppose the latter could
happen, given that some students insist on dwelling in a fog of extreme
cluelessness. However, I can't imagine breathing a sigh of relief that LHR
is there to help this student.
> Now imagine a similar limit only with a much more
> hairy numerator that is identically zero but is not
> immediately recognizably so. As above, L'Hospital's
> rule might similarly simply reveal the correct result.
You must verify the hairy numerator -> 0 to use LHR. If it's identically 0
but is unrecognizable as such, how will you know it -> 0?
> The same remark apply to machines as well as humans.
> Computer algebra systems may encounter such limits.
>
> Further, suppose that the rule didn't apply in the
> case the numerator is identically zero. Then before
> applying the rule one would be required to verify
> also that the numerator was not identically zero.
> But the zero equivalence problem is undecidable
> in general (even so for most any nontrivial class
> of functions one encounters in calculus, see [1]).
I have no idea what you're talking about here.
>
> So it's a good thing the rule applies in this case.
It's a good thing the rule applies because the theorem asserts that it
applies and we do not wish mathematics to be contradictory. What is not a
good thing is to use LHR to discover the value of lim_s->0 0/s^2. Because
that would be ridiculous. (It is not ridiculous to check that LHR gives us
the right answer, but that's a different matter.)
Daniel W. Johnson
> First, to apply LHR, you must verify the
> numerator -> 0; how will you do that without knowing sin^2 + cos^2 = 1?
By knowing that sin 0 = 0 and cos 0 = 1.
--
Daniel W. Johnson
pano...@iquest.net
http://members.iquest.net/~panoptes/
039 53 36 N / 086 11 55 W
Denis Feldmann
> Robin Chapman <r...@ivorynospamtower.freeserve.co.uk> wrote:
>> lihuijun <lihu...@hotmail.com> wrote: (edited)
>>>
>>> Can we use L/hopital rule here?
>>>
>>> 0/x as x-->0
>>>
>>> 0/x^2 as x-->0
>>>
>>> If so, then the ans is 0. Please advise
>>
>> L'hopital's rule is rarely useful, but this must be
>> the most pointless application of it ever devised!
>
> Before rushing to judgment, consider the following
>
> 2 2
> sin (x) + cos (x) - 1
> lim ---------------------
> x->0 x^2
>
>
> The expression equals 0/x^2, so has the above form.
> Applying L'Hopital's rule reduces it to the trivial
> limit of 0/(2x) and thus yields the correct result
> _without_ requiring the specialized trig knowledge
> that the initial numerator is identically zero.
But then, how do you know that it is of the required form "0/0"? The same
trick would work for
2 2
sin (x) + cos (x) - 2
lim ---------------------
x->0 x^2
... except that it doesn't
David C. Ullrich
<wadera...@comcast.remove13.net> wrote:
>In article <y8zd60o...@nestle.csail.mit.edu>,
> Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
>
>> Before rushing to judgment, consider the following
>>
>> 2 2
>> sin (x) + cos (x) - 1
>> lim ---------------------
>> x->0 x^2
>>
>>
>> The expression equals 0/x^2, so has the above form.
>> Applying L'Hopital's rule reduces it to the trivial
>> limit of 0/(2x) and thus yields the correct result
>> _without_ requiring the specialized trig knowledge
>> that the initial numerator is identically zero.
>
>That's dubious for two reasons. First, to apply LHR, you must verify the
>numerator -> 0; how will you do that without knowing sin^2 + cos^2 = 1?
>Second, we're to believe someone has forgotten the most basic of trig
>identities, sort of like forgetting your own name, and yet knows how to
>apply the chain rule and remember the derivatives of sin and cos correctly?
Third, if we know the derivatives of sin and cos (and their values at
the origin) then the fact that sin^2 + cos^2 = 1 follows trivially.
************************
David C. Ullrich
Dave Seaman
> The World Wide Wade <wadera...@comcast.remove13.net> wrote:
>> First, to apply LHR, you must verify the
>> numerator -> 0; how will you do that without knowing sin^2 + cos^2 = 1?
> By knowing that sin 0 = 0 and cos 0 = 1.
Only if the student can also prove that sin^2 x + cos^2 x - 1 is
continuous at x = 0.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>
Bill Dubuque
>>
>> Before rushing to judgment, consider the following
>>
>> 2 2
>> sin (x) + cos (x) - 1
>> lim ---------------------
>> x->0 x^2
>>
>>
>> The expression equals 0/x^2, so has the above form.
>> Applying L'Hopital's rule reduces it to the trivial
>> limit of 0/(2x) and thus yields the correct result
>> that the initial numerator is identically zero.
>
> That's dubious for two reasons. First, to apply LHR, you must verify the
> numerator -> 0; how will you do that without knowing sin^2 + cos^2 = 1?
> Second, we're to believe someone has forgotten the most basic of trig
> identities, sort of like forgetting your own name, and yet knows how to
> apply the chain rule and remember the derivatives of sin and cos correctly?
You're missing the point. Since zero equivalence is undecidable,
in some cases the numerator f(x) will be identically zero without
you having knowledge of such; but you may know that f(0) = 0, etc.
Then lim f(x)/x^2 would invoke the special case of L'Hospital
mentioned by the o.p. and it might actually prove useful in the
sense that, after differentiation, the new limit is "simpler",
As I said in the next paragraph, one is supposed to imagine
a much hairier example where this occurs. Instead you are
focusing on the details of this particular example, which
perhaps is far too specialized to abstract the essence from.
>> Now imagine a similar limit only with a much more
>> hairy numerator that is identically zero but is not
>> immediately recognizably so. As above, L'Hospital's
>> rule might similarly simply reveal the correct result.
>
> You must verify the hairy numerator -> 0 to use LHR. If it's identically 0
> but is unrecognizable as such, how will you know it -> 0?
I assume that f(0)=0 is known. Often it's much simpler
to prove that f(0)=0 versus f(x)=0 identically. In the
case at hand it requires only: sin 0 = 0, cos 0 = 1.
>> The same remark apply to machines as well as humans.
>> Computer algebra systems may encounter such limits.
>>
>> Further, suppose that the rule didn't apply in the
>> case the numerator is identically zero. Then before
>> applying the rule one would be required to verify
>> also that the numerator was not identically zero.
>> But the zero equivalence problem is undecidable
>> in general (even so for most any nontrivial class
>> of functions one encounters in calculus, see [1]).
>
> I have no idea what you're talking about here.
Is it clearer now? If not, please feel free to elaborate.
>> So it's a good thing the rule applies in this case.
>
> It's a good thing the rule applies because the theorem asserts that it
> applies and we do not wish mathematics to be contradictory. What is not a
> good thing is to use LHR to discover the value of lim_x->0 0/x^2. Because
> that would be ridiculous. (It is not ridiculous to check that LHR gives us
> the right answer, but that's a different matter.)
See above.
--Bill Dubuque
Bill Dubuque
>Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
>> Robin Chapman <r...@ivorynospamtower.freeserve.co.uk> wrote:
>>> lihuijun <lihu...@hotmail.com> wrote: (edited)
>>>>
>>>> Can we use L/hopital rule here?
>>>>
>>>> 0/x as x-->0
>>>>
>>>> 0/x^2 as x-->0
>>>>
>>>> If so, then the ans is 0. Please advise
>>>
>>> L'hopital's rule is rarely useful, but this must be
>>> the most pointless application of it ever devised!
>>
>> Before rushing to judgment, consider the following
>>
>> 2 2
>> sin (x) + cos (x) - 1
>> lim ---------------------
>> x->0 x^2
>>
>>
>> The expression equals 0/x^2, so has the above form.
>> Applying L'Hopital's rule reduces it to the trivial
>> limit of 0/(2x) and thus yields the correct result
>> that the initial numerator is identically zero.
>
> But then, how do you know that it is of the required form "0/0"?
> The same trick would work for
>
> 2 2
> sin (x) + cos (x) - 2
> lim ---------------------
> x->0 x^2
>
> ... except that it doesn't
You misread what I wrote. I said that L'Hospital's rule
yields the correct result without employing any special
knowledge that the numerator f(x) is _identically_ zero.
Obviously, for the rule to apply, one stills needs to
check that f(0) = 0. But this is often much simpler
than proving that f(x) is _identically_ zero.
For further discussion see my reply to The World Wide Wade.
--Bill Dubuque
Christopher
> In article <y8zd60o...@nestle.csail.mit.edu>,
> Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
>
> > Before rushing to judgment, consider the following
> >
> > 2 2
> > sin (x) + cos (x) - 1
> > lim ---------------------
> > x->0 x^2
> >
> >
> > The expression equals 0/x^2, so has the above form.
> > Applying L'Hopital's rule reduces it to the trivial
> > limit of 0/(2x) and thus yields the correct result
> > _without_ requiring the specialized trig knowledge
> > that the initial numerator is identically zero.
>
> That's dubious for two reasons. First, to apply LHR, you must verify the
> numerator -> 0; how will you do that without knowing sin^2 + cos^2 = 1?
That's easy. You know sin and cos are continuous and you know their
values at 0. What kind of calculus did you teach, anyway?
> Second, we're to believe someone has forgotten the most basic of trig
> identities, sort of like forgetting your own name, and yet knows how to
> apply the chain rule and remember the derivatives of sin and cos correctly?
He was obviously just using a simple example to make the point. See
below if you insist on a more complicated one.
> > Now imagine a similar limit only with a much more
> > hairy numerator that is identically zero but is not
> > immediately recognizably so. As above, L'Hospital's
> > rule might similarly simply reveal the correct result.
>
> You must verify the hairy numerator -> 0 to use LHR. If it's identically 0
> but is unrecognizable as such, how will you know it -> 0?
Consider instead:
(4(cos(x)-cos(3x))/(1+sec(x)) - 8((1-cos(x))/sec(x))^2)^2 -
(1-cos(4x))^2
It takes about 10 seconds to verify that it's continuous at 0, and has
a value of 0 there. Now, I don't know if taking the derivative of this
expression and evaluating it at x=0 is easier than showing it's
identically equal to 0, but I would believe someone if they told me it
was easier for them.
Dae-jung Yoo
Obviously, limit_{s-->0+} 0/s = 0 , because 0/s is zero for every positive
number s.
0 --> 0 as s-->0+.
s --> 0 as s-->0+.
d/ds (0) =0.
d/ds (s) = 1.
So L'hopitals rule also gives
limit 0/s = limit 0/1 = 0.
"lihuijun" <lihu...@hotmail.com> wrote in message
news:200409132059...@proapp.mathforum.org...
The World Wide Wade
Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
> The World Wide Wade <wadera...@comcast.remove13.net> wrote:
> >Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
> >>
> >> Before rushing to judgment, consider the following
> >>
> >> 2 2
> >> sin (x) + cos (x) - 1
> >> lim ---------------------
> >> x->0 x^2
> >>
> >>
> >> The expression equals 0/x^2, so has the above form.
> >> Applying L'Hopital's rule reduces it to the trivial
> >> limit of 0/(2x) and thus yields the correct result
> >> without requiring the specialized trig knowledge
> >> that the initial numerator is identically zero.
> >
> > That's dubious for two reasons. First, to apply LHR, you must verify the
> > numerator -> 0; how will you do that without knowing sin^2 + cos^2 = 1?
> > Second, we're to believe someone has forgotten the most basic of trig
> > identities, sort of like forgetting your own name, and yet knows how to
> > apply the chain rule and remember the derivatives of sin and cos correctly?
>
> You're missing the point. Since zero equivalence is undecidable,
> in some cases the numerator f(x) will be identically zero without
> you having knowledge of such; but you may know that f(0) = 0, etc.
> Then lim f(x)/x^2 would invoke the special case of L'Hospital
> mentioned by the o.p. and it might actually prove useful in the
> sense that, after differentiation, the new limit is "simpler",
Actually my original point was that using LHR to evaluate lim_s->0 0/s^2 is
silly. You termed this "rushing to judgment", which seems strange because
you wrote "the trivial limit of 0/(2x)". Exactly.
You seem to have other points to make, one of which is that a hairy
numerator can be identically 0 without this being obvious. You are clearly
right, and I don't know why I raised any objection to it. But that's a
different point altogether. You also say it's a good thing LHR applies in
trivial cases like 0/s^2, otherwise ... And my answer to that is still the
same: Sure it's "good" that it applies, but it's bad to apply it: using it
to evaluate lim_s->0 0/s^2 (as the OP suggested) is silly. That's my point
and I'm sticking to it.
The World Wide Wade
ni...@fas.harvard.edu (Christopher) wrote:
> The World Wide Wade wrote ...
> > In article <y8zd60o...@nestle.csail.mit.edu>,
> > Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
> >
> > > Before rushing to judgment, consider the following
> > >
> > > 2 2
> > > sin (x) + cos (x) - 1
> > > lim ---------------------
> > > x->0 x^2
> > >
> > >
> > > The expression equals 0/x^2, so has the above form.
> > > Applying L'Hopital's rule reduces it to the trivial
> > > limit of 0/(2x) and thus yields the correct result
> > > _without_ requiring the specialized trig knowledge
> > > that the initial numerator is identically zero.
> >
> > That's dubious for two reasons. First, to apply LHR, you must verify the
> > numerator -> 0; how will you do that without knowing sin^2 + cos^2 = 1?
>
> That's easy. You know sin and cos are continuous and you know their
> values at 0. What kind of calculus did you teach, anyway?
I doubt one "knows" the values of sin and cos at 0, and that these are
continuous functions, without knowing sin^2 + cos^2 = 1. But yes, you could
remember those facts in isolation.
> > Second, we're to believe someone has forgotten the most basic of trig
> > identities, sort of like forgetting your own name, and yet knows how to
> > apply the chain rule and remember the derivatives of sin and cos correctly?
>
> He was obviously just using a simple example to make the point. See
> below if you insist on a more complicated one.
>
> > > Now imagine a similar limit only with a much more
> > > hairy numerator that is identically zero but is not
> > > immediately recognizably so. As above, L'Hospital's
> > > rule might similarly simply reveal the correct result.
> >
> > You must verify the hairy numerator -> 0 to use LHR. If it's identically 0
> > but is unrecognizable as such, how will you know it -> 0?
>
> Consider instead:
>
> (4(cos(x)-cos(3x))/(1+sec(x)) - 8((1-cos(x))/sec(x))^2)^2 -
> (1-cos(4x))^2
>
> It takes about 10 seconds to verify that it's continuous at 0, and has
> a value of 0 there. Now, I don't know if taking the derivative of this
> expression and evaluating it at x=0 is easier than showing it's
> identically equal to 0, but I would believe someone if they told me it
> was easier for them.
True, and as I wrote to Bill, I shouldn't have gotten involved with that
issue, because the only point I was trying to make is that it is silly to
use LHR to evaluate lim_s->0 0/s^2. Which it is.
Lee Rudolph
> values at 0. What kind of calculus did you teach, anyway?
The functional calculus, perhaps?
Lee Rudolph (whereas, if my University-Wide Teaching Evaluation forms
can be believed, I apparently have a lock on the dysfunctional calculus)
Bill Dubuque
> is silly. You termed this "rushing to judgment", which seems strange
> because you wrote "the trivial limit of 0/(2x)". Exactly.
There is some implicit notational overloading here that may be
the source of some confusion. When one says that a limit is of
a certain "form", e.g. lim f/g has form lim 0/x^2, one usually
denotes the following _function_ equalities: f = 0, g = x^2.
Here f = 0 is interpreted semantically as a function equality,
vs. syntactically, i.e. that the _expression_ f is literally "0".
This is as it should be since one takes the limit of functions,
not expressions. Otoh, when I said above that it "reduces to the
trivial limit of 0/(2x)" the term "0/2x" is not meant to denote
a limit "form" but instead a function. Indeed I referred to the
"limit of f/g" not a "limit of the form f/g". Thus I agree that
it's silly to apply L'Hopital's rule to lim 0/x^2 but it's not
so silly to apply it to a limit of the form 0/x^2, as x -> 0,
for reasons that I've already described at length in prior posts.
By the way, the tension between semantic vs. syntactic equality
is a common source of confusion in programming languages that
employ pattern matching. In some contexts one desires the match
to be based on literal equality and in other cases one desires
a more semantic interpretation. In less precise forums such as
mathematical prose one usually depends on the readers so-called
mathematical maturity to intelligently resolve ambiguities.
> You seem to have other points to make, one of which is that a hairy
> numerator can be identically 0 without this being obvious. You are
> clearly right, and I don't know why I raised any objection to it.
> But that's a different point altogether. You also say it's a good
> thing LHR applies in trivial cases like 0/x^2, otherwise ... And my
> answer to that is still the same: Sure it's "good" that it applies,
> but it's bad to apply it: using it to evaluate lim_x->0 0/x^2
> (as the OP suggested) is silly. That's my point and I'm sticking to it.
Even for the limited class of elementary functions encountered in a
a first course in calculus, it's undecidable to decide if a function
is zero or not. So if I give you a limit to compute via LHR you can't
tell in general if you're applying LHR to a limit of the form 0/x^2.
Don't worry, as much as you dislike it, you can't even tell if you
are doing it, so you shouldn't be too pained sticking to your point.
But if your pain becomes too severe, try another form of L'Hospital.
There's a special one devoted to cases where you have flipped your
lid or numerator, which no doubt you'll need to do in order to deal
with the depressingly undecidable problem of zero-equivalence.
Flippantly,
--Bill Dubuque
AaronB
> In article <y8zd60o...@nestle.csail.mit.edu>,
> Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
>
> > Before rushing to judgment, consider the following
> >
> > 2 2
> > sin (x) + cos (x) - 1
> > lim ---------------------
> > x->0 x^2
> >
> >
> > The expression equals 0/x^2, so has the above form.
> > Applying L'Hopital's rule reduces it to the trivial
> > limit of 0/(2x) and thus yields the correct result
> > _without_ requiring the specialized trig knowledge
> > that the initial numerator is identically zero.
>
> That's dubious for two reasons. First, to apply LHR, you must verify the
> numerator -> 0; how will you do that without knowing sin^2 + cos^2 = 1?
> Second, we're to believe someone has forgotten the most basic of trig
> identities, sort of like forgetting your own name, and yet knows how to
> apply the chain rule and remember the derivatives of sin and cos correctly?
> Well, having taught enough calculus by now, I suppose the latter could
> happen, given that some students insist on dwelling in a fog of extreme
> cluelessness. However, I can't imagine breathing a sigh of relief that LHR
> is there to help this student.
[snip]
I think that this is simply a poor excuse to use L'Hopital. Instead
consider the following examples:
y = lim x-> 0 (sinx)/x
y = lim x-> 0 (e^x - 1)/x
It is extremely easy to show that both of these equations are of the
form 0/0, and taking L'Hopital's rule of both is trivial. Solving
these without L'Hopital's rule, on the other hand, is not immediately
obvious.
A.
The World Wide Wade
amino_...@hotmail.com (AaronB) wrote:
> I think that this is simply a poor excuse to use L'Hopital. Instead
> consider the following examples:
>
> y = lim x-> 0 (sinx)/x
> y = lim x-> 0 (e^x - 1)/x
>
> It is extremely easy to show that both of these equations are of the
> form 0/0, and taking L'Hopital's rule of both is trivial. Solving
> these without L'Hopital's rule, on the other hand, is not immediately
> obvious.
Well in fact they are obvious: The first is by definition the derivative of
sin(x) at 0, which is 1. The second is the derivative of e^x at 0, which is
1. You'll find that many exercises on LHR can be better done, with greater
insight, by realizing that the limit at hand is just a derivative
(sometimes in disguise).
Ronald Bruck
And why not? After all, it's clear that the OP didn't have questions
of formal undecidability of recognizing whether the numerator is
identically zero. He wrote a gratuitous "(s positive)", for heaven's
sake! He's not only a calculus student, he's a BAD calculus student!
--Ron Bruck