View Full Version : l'Hopital's Rule

ygfperson

01-16-2003, 06:18 PM

I'm having a discussion (read: argument) with my calculus teacher about this.

lim cos(x) - .5

---------

h->PI/6 PI/6 - x

I say that if you evaluate this limit it turns out undefined. She says that since the expression results in division by zero, l'Hopital's rule can be used.

lim sin(x)

---------

h->PI/6 1

and this results in 0.5

Who is right? I think I am, but I want a general concensus here so I don't keep fighting a losing battle here.

Polymorphic OOP

01-16-2003, 06:30 PM

your teacher is corect, however you are confused.

The limit has the potential of existing not because the denominator is 0, but because BOTH the numerator AND the denominator are 0.

In other circumsatnces (IE without trigonometric functions) you can do things like factoring and cancelation (logic being that 0/0 means both the numerator and the denominator have a common factor) or use other methods.

Cshot

01-16-2003, 06:44 PM

I think you're both wrong.

Shouldn't it be x->pi/6 instead of h->pi/6 :D

Anyways, shouldn't the answer be infinity since the numerator isn't zero at x=pi/6?

alpha

01-16-2003, 06:55 PM

Cshot, I'm not sure because you would end up with .366/0, and that isn't possible. also, yes, it should be x-> pi/6, probably just a typo.

Yes, your teacher is right though, hopefully not with this problem. L'Hopitals Rule can be used with these following inderterminate forms:

0/0

infinity/infinity

infiniti - infinity

i think there is more, i just can't remember them off the top of my head right now, and my calc notes are at school.

Cshot

01-16-2003, 07:00 PM

>> Cshot, I'm not sure because you would end up with .366/0, and that isn't possible.

You're right that .366/0 is undefined. But this is a limit problem. As x->pi/6, the limit goes to infinity.

And yea, I was just joking about the h->pi/6 part :)

alpha

01-16-2003, 07:06 PM

this particular limit does not classify as being able to use L'Hopital's Rule, this limit is undefined. I ran this through my calculator, and it gave me an answer of undefined (wonderful uses of the TI-89). If cos(x) - .5 as x -> pi/6 evaluated to zero, then i can see this working. as i hope this particular problem is not the one being discussed and does have typos.

Cshot

01-16-2003, 07:09 PM

I agree with you alpha that this isn't a l'hopital problem. Just solve it like any other limit problem. The numerator goes to a constant while the denominator approaches 0, therefore the entire limit approaches infinity.

If the problem were to say what is (cos(x)-0.5)/(pi/6 - x) at x=pi/6, then the answer would be undefined.

Polymorphic OOP

01-16-2003, 07:09 PM

ha, you're right, i didn't notice the numerator wasn't 0. You'd think that after I specifically said the numerator had to be 0 that would have checked if it was in this example :p

cos( pi/3 ) is 1/2 NOT cos( pi/6 )

alpha

01-16-2003, 07:15 PM

Cshot, it still doesn't make sense. The calculator does do limits and it says the answer is undefined. if the denominator approached 1, then i could see how it approaches a constant; but it approaches 0, and once pi/6 is plugged in just like any limit problem, it is evaluated with the values, and .366/0 is undefined.

I'm still in high school Calc BC, so maybe I'm missing something here...but I don't think I am.

edit: Actually, come to think of it, if the limit was as x approaches infinity, then the limit would evaluate to infinity...

golfinguy4

01-16-2003, 07:47 PM

My Mathmatical Theory (which you people are free to shoot down and you probably will):

I think that inf/inf should equal 1. x/x=1 for non-zero integers. Even the lim as h->inf of h/h=1. Therefore, since infinity is just a very very big number, shouldn't the same be true? This would make the integration of divergent functions very easy. For instance, with this idea, int from -2 to 2 of 1/x dx would equal 0.

alpha

01-16-2003, 08:12 PM

Originally posted by golfinguy4

My Mathmatical Theory (which you people are free to shoot down and you probably will):

I think that inf/inf should equal 1. x/x=1 for non-zero integers. Even the lim as h->inf of h/h=1. Therefore, since infinity is just a very very big number, shouldn't the same be true? This would make the integration of divergent functions very easy. For instance, with this idea, int from -2 to 2 of 1/x dx would equal 0.

you can't take the ln of negative numbers, so the integral from -2 to 2 of 1/x dx would not equal zero. and inf/inf is indeterminate, because infinity is a concept. inf/inf could be 1, could be 5, etc.

golfinguy4

01-16-2003, 08:40 PM

Why do you mention ln? Yes, ln is the anti-derivative of 1/x. However, if you read the FToC carefully, you will see that the antidifferentiation rule is only guaranteed to work if f is continuous along the interval.

Cshot

01-16-2003, 09:37 PM

>> I think that inf/inf should equal 1. x/x=1 for non-zero integers.

Nah, you can't apply normal math operations with infinity. I think they're called indeterminants and normal math rules don't apply.

Hmm, alpha you may be right.

I looked it up here:

http://www.alltel.net/~okrebs/page191.html

It says:

If lim f(x) != 0 and lim g(x) = 0 then one of the following is true:

a) lim f(x)/g(x) = inf or lim f(x)/g(x) = -inf

b) lim f(x)/g(x) doesn't exist

Seems like if you approach pi/6 from the right, you'll approach negative infinity. If you approach pi/6 from the left, you'll approach positive infinity. Therefore, the limit doesn't exist for this case.

I think you are right. This is because you can only use lhr

when 0/0, inf/inf or -inf/inf.

You might want to show this to

your teacher

1/h as h->0 is certainly undefined. What follows gives

a similar expression

(cos(x) - .5) / (pi/6 -x) as x->pi/6 is equivalent to

(cos(x + h) - .5)/h as h->0 where x is fixed to be pi/6.

This is the same as

(cos(x + h) - cos(x) + (sqrt(3) - 1)/2) / h as h->0 which comes to

-sin(x) + (sqrt(3) - 1)/2h as h->0

I think that inf/inf should equal 1. x/x=1 for non-zero integers. Even the lim as h->inf of h/h=1. Therefore, since infinity is just a very very big number, shouldn't the same be true? This would make the integration of divergent functions very easy. For instance, with this idea, int from -2 to 2 of 1/x dx would equal 0.

If inf/inf equals 1 then x / (2x) as x->inf would be 1.

Wouldn't you just integrate 1/x and get ln |x| from -2 to 2 and

so then you have 0. integral(1/x) is not ln x.

ygfperson

01-16-2003, 10:17 PM

Originally posted by Cshot

>> Cshot, I'm not sure because you would end up with .366/0, and that isn't possible.

You're right that .366/0 is undefined. But this is a limit problem. As x->pi/6, the limit goes to infinity.

And yea, I was just joking about the h->pi/6 part :)

yeah, that is a typo on my part. it's supposed to be x -> pi/6. but the rest is the way it is on the test.

one related question: when the limit of a function is plus or minus infinity, isn't the answer still undefined? i thought that the use of infinity here was just showing one type of undefined answer, acting as notation and not as a number.

It's undefined in the sense that it's limit is not a real number.

But there is a formal definition of when the limit

is +-infinity.

alpha

01-16-2003, 10:53 PM

Cshot, that is why I thought it didn't exist, because left-hand limit != right-hand limit.

golfinguy:

S[-2, 2] 1/x dx

= ln |x| ] [-2, 2]

= ln (2) - ln (-2)

ln of negative numbers can't happen. yes you can integrate 1/x as ln x.

the S is supposed to be integral, closest thing. [-2, 2] are the limits.

I was wrong about the integral of 1/x.

You have to integrate 1/x as two seperate integrals

from [-2, 0) to (0, 2] But then integeral of 1/x

is ln|x| which is undefined at x = 0.

Sang-drax

01-17-2003, 06:48 AM

I'd say it's undefined, and Mathcad agrees.

ygfperson

01-17-2003, 03:23 PM

argh!! I explained it to my math teacher (again) and she still thinks she's right...

What can convince a person beyond mathematical proof?

golfinguy4

01-17-2003, 03:26 PM

Originally posted by Nick

If inf/inf equals 1 then x / (2x) as x->inf would be 1.

Wouldn't you just integrate 1/x and get ln |x| from -2 to 2 and

so then you have 0. integral(1/x) is not ln x.

That's not what I mean.... I meant that as a last resort, inf./inf. should be evaluated as 1.

golfinguy4

01-17-2003, 03:30 PM

Originally posted by alpha

Cshot, that is why I thought it didn't exist, because left-hand limit != right-hand limit.

golfinguy:

S[-2, 2] 1/x dx

= ln |x| ] [-2, 2]

= ln (2) - ln (-2)

ln of negative numbers can't happen. yes you can integrate 1/x as ln x.

the S is supposed to be integral, closest thing. [-2, 2] are the limits.

However, you fail to realize that you can't integrate a non-continuous function like that. The method that you are stating is from the Fundamental Theorem of Calculus. However, the FToC says that the anti-differentiation method only works when the function is continuous on the interval. As you can see, 1/x is not continuous from -2,2.

BTW, I know that I am wrong on this.... It is just that I think that this should be the way things work.

alpha

01-17-2003, 04:37 PM

yes, i know tha FToC only applies to continuous functions. I realized it later, thanks.

*ClownPimp*

01-17-2003, 04:51 PM

the fundamental theorem may not apply directly, but you can split the function into two continuous pieces and use improper integrals (lim b->c Int [...] ).

>I meant that as a last resort, inf./inf. should be evaluated as 1.

Not necessarily. It depends on how fast one goes to infinity compared to the other...

lim x->inf. (exp(x)/x) certainly doesnt equal 1. L'H. rules basically determines which one goes to inf./zero faster

What can convince a person beyond mathematical proof?

You should be able to prove it goes to infinity.

(cos(x) - .5) / (pi/6 -x) as x->pi/6 cos(pi/6) ~= .866

It's somewhat sloppy but could be justified this limit

is greater than

.1 / h as h->0 = (.1) * 1/h as h->0

1/h as h->0 is infinity if your teacher does not

accept this as a informal proof then she should

not be teaching calculus.

To be fair to your teacher I think she wanted the problem

to be

(sin(x) - .5) / (pi/6 -x) as x->pi/6

ygfperson

01-17-2003, 08:20 PM

*rant warning*

Originally posted by Nick

To be fair to your teacher I think she wanted the problem

to be

(sin(x) - .5) / (pi/6 -x) as x->pi/6

i'm guessing that too... she probably got the problem from a book, and wrote it down wrong. since close to everyone in class just used l'hopital's rule without checking to see if it would fit, everyone else got the same answer. I can see how she doubts me.

but still... it's really annoying to get points taken off for someone else's mistake. (although she gave me 4 out of 8 of the points for effort).

I'm taking the AP calculus midyear exam tuesday, and she said we could discuss it further after the exam. If I can't prove I'm right by or at that time, that grade will be submitted into the school's computers and beyond change.

So far I've given her the proof based on the weak proof of l'hopital's theorem, based on the limit derivatives are made of. (it also works for the strong proof, based on the mean value theorem.) i've showed her where in my book it says that it needs a 0/0 form in order to be used. I've drawn up a graph of it and showed that there's an asymtote at x=pi/6, where the right handed and left handed limits don't touch. I've done it on the TI-89 calculator.

she believes that i typed it in wrong in the ti-89 calculator, or i forgot to surround the whole expression with parenthesis. she believes that when the book says that it only works with a inf/inf or 0/0 indeterminate form, division by 0 is counted as a 0/0 form. she's talked with a group of calculus teachers, and some calculus teachers in school, and they agree with her. i can only think that her explanation of the problem was wrong or misleading.

I realize I have plenty of proof to back up my view. All of you agree with me, at least in the end. It works out graphically, and it proves itself mathematically. Still...

How do you convince a person who's determined she's right? She's not necessarily closed-minded to the fact that she's wrong in this area. I think she's always done it this way, and hasn't dealt with enough different l'Hopital problems to see that she's wrong.

The only thing I can think of is to present as much evidence as possible to her in that time. Not just mathematical proof (although that's necessary) but some opinions from real people. (Not that you all aren't real... you know what I mean. ;)) If I'm only showing her the math, she may be convinced I have a flaw in my reasoning somewhere.

//edit: she pointed out a similar problem solved using l'hopital's:

lim cos x - 0.5

-----------

x->pi/3 x - pi/3

However this problem evaluates to 0/0, which makes it solvable by L'Hopital's... right?

I have this weird premonition that I'll go up to her about this (yet again) and she'll respond by saying that there was a typo in that problem that I didn't see and the rest of the class did and corrected. Maybe I'm just nervous...

joshdick

01-17-2003, 09:00 PM

To whomever thought infinity/infinity = 0,

Infinity is not a number; it is a concept. You simply cannot treat it like a number. Think about some infinite sets. (-inf, inf) is the set of all real numbers. That's about the biggest infinity can get. However, (0,inf) is the set of all positive reals and that, too is and infinite amount of numbers. Just think about all of the numbers between 0 and 1. There are an infinite amount of numbers just between 0 and 1.

See how some infinities seem more than others? That's why they're indeterminate. Your suggestion that inf/inf should be evaluated to 1 as a last resort holds no water. Any number would be just as good of a guess and just as bad of an idea as 1.

And to all out there that think that the natural log a negative number is undefined,

that's only if you are restricting your math to real numbers. I really like imaginary and complex numbers. Euler came up with a relationship that allows the ln of a negative number to be evaluated.

e^(pi * i) + 1 = 0

I think that's what it is. Just manipulate that identity a bit and you can find the natural logs of negative numbers. You can do this even simpler by setting your TI-83 to complex number mode. Then, you'll be able to evaluate lns of negative numbers.

I wish my calculus class allowed for imaginary and complex numbers. My calc teacher said that those numbers simply aren't allowed in any of the work we do. That was a real drag after having learned DeMoivre's Theorem in Pre-Calc. I'm looking forward to a time in college when complex numbers will be allowed.

alpha

01-18-2003, 08:22 AM

ygfperson, I'm surprised she thought that you typed in wrong on the TI89. Is it set to pretty print? If it is, it should look the exact same way as the problem. I'm surprised she is a calc teacher if she believes that 2/0 for example could be indeterminate. 0/0 is indeterminate, and any other number/0 is undefined. The rest of the class is stupid, if they don't check to see if L'Hopital's Rule fits. At least on the AP exam, you would get the problem right and the rest of the class wouldn't (at least if that problem was on the AP).

ygfperson

02-11-2003, 09:31 PM

Excuse me for bumping... I'm just adding new info to the thread.

My teacher gave me a name of a college professor (who taught my math teacher) to talk to about the problem. I called up, and she agreed with my teacher. But then I told her cos(pi/6) was sqrt(3)/2, not 1/2. She commented on how this all was a while ago and she said I was right about the cos(pi/6) thing. So we discuss the problem, and she comes to the same conclusion we've all reached: The limit is to infinity.

I told this to my math teacher, and she looked suprised, and asked me to explain the problem with her. So that's what I'm going to do tomorrow morning, during my study. (No homework to do anyway... :)).

Here's what I'm thinking of saying:

1) Mean Value theorem: f'(c) = (f(b) - f(a))/(b-a)

where a and b are points, and c is in between.

2) Definition of a derivative: f'(c) = lim x->a (f(x)-f(a))/(x-a)

Says the same thing as mean value theorem, but less rigorous.

3) Given two functions, f(x) and g(x):

f'(c)/g'(c) == (f(b) - f(a))/(g(b) - g(a))

The (b-a) expression cancels out.

4) Take the limit as b -> a

lim b->a f'(c)/g'(c) == lim b->a ...

5) Assume that f(a) = g(a) = 0

lim b->a f'(c)/g'(c) == (f(b)/g(b))

6) Due to the Squeeze or Sandwich theorem:

lim b->a f'(b)/g'(b) == f(b)/g(b)

7) if f(a) != 0

lim b->a f'(c)/g'(c) == lim b->a (f(b) - f(a))/g(b)

8) So L'Hopital doesn't work in that case.

Anyone see any holes in that?

*ClownPimp*

02-12-2003, 09:16 AM

What exactly are you trying to prove?

on (7) where you have lim b->a f'(c)/g'(c) == lim b->a (f(b) - f(a))/g(b), your forgetting that in the limit as b->a, f(b) - f(a) == 0, and since g(b) = 0 (since it was "approaching" g(a), which == 0), you still have an indeterminate case of 0/0, so LH's rule still applies

ygfperson

02-12-2003, 03:34 PM

your forgetting that in the limit as b->a, f(b) - f(a) == 0

I thought about that, too, but I think that's wrong. I'm not sure why, but I'll take another stab at it.

To make things easier, I'll call it:

lim x->x0 f'(c)/g'(c) == lim x->x0 [f(x)-f(x0)]/[g(x)-g(x0)]

This gets into very specific facts about limits. x != x0. It's a given that f(x0) = g(x0) = 0. f(x) is not exactly f(x0).

Limits at their center are based on continuous functions.

Let's say that lim x->h (x+1)/(x+2)

The difference between the limit and the number is called epsilon.

x - epsilon = h

for that epsilon, there is a value 'delta' which is the difference between the function and limit of the function.

f(x) - delta = f(h)

As epsilon gets very small, delta gets very small too, due to the continuous nature of functions. x != h, but x can get very close to h.

Getting back to my problem, f(x) - f(x0) is close to zero, but f(x0) is zero. That takes presedence. No matter how close f(x) is to f(x0), it will never be 0.

Can someone else who's been through this elaborate on this, too? I think I'm wrong somewhere...

One more thing: I've finally convinced her I'm right. It turns out all along she thought cos(30) was .5

alpha

02-12-2003, 04:44 PM

Originally posted by ygfperson

One more thing: I've finally convinced her I'm right. It turns out all along she thought cos(30) was .5

trig facts coming back to haunt you...lol...

i need to brush up on trig too though.

well, that explains it. that should be enough, that you are not taking the limit of an indeterminate form, so l'hopitals rule doesn't apply.

*ClownPimp*

02-12-2003, 05:55 PM

>lim x->x0 f'(c)/g'(c) == lim x->x0 [f(x)-f(x0)]/[g(x)-g(x0)]

>

>This gets into very specific facts about limits. x != x0. It's a given >that f(x0) = g(x0) = 0. f(x) is not exactly f(x0).

That is true, but my point still stands. lim b->a[ (f(b) - f(a))/g(b) ] is an indeterminate form and cannot be evaluated directly. If you were to try to evaluate it directly, what value is b....?

*ClownPimp*

02-12-2003, 06:13 PM

Also,

>5) Assume that f(a) = g(a) = 0

>lim b->a f'(c)/g'(c) == (f(b)/g(b))

lim b->a f'(c)/g'(c) = lim b->a [ (f(b) - f(a))/(g(b) - g(a)) ] =

lim b->a [ (f(b) - f(a))/(g(b) - g(a)) ] = lim b->a [ (f(b) - 0)/(g(b) - 0) ] != f(b)/g(b)

Once you take the limit of something you have to evaluate the limit. b is not a constant value here so you cant just say lim b->a f(b) = f(b) because b outside of the lim has no meaning

ygfperson

02-12-2003, 09:18 PM

d'oh... #5 was a typo. Or I just forgot to include the limit.

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