l'Hopital's Rule

This is a discussion on l'Hopital's Rule within the A Brief History of Cprogramming.com forums, part of the Community Boards category; Originally posted by Cshot >> Cshot, I'm not sure because you would end up with .366/0, and that isn't possible. ...

  1. #16
    Just because ygfperson's Avatar
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    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.

  2. #17
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    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.

  3. #18
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    Cshot, that is why I thought it didn't exist, because left-hand limit != right-hand limit.

    golfinguy:
    Code:
    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.

  4. #19
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    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.

  5. #20
    S Sang-drax's Avatar
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    I'd say it's undefined, and Mathcad agrees.
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    Last edited by Sang-drax : Tomorrow at 02:21 AM. Reason: Time travelling

  6. #21
    Just because ygfperson's Avatar
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    argh!! I explained it to my math teacher (again) and she still thinks she's right...

    What can convince a person beyond mathematical proof?

  7. #22
    ¡Amo fútbol!
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    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.

  8. #23
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    Originally posted by alpha
    Cshot, that is why I thought it didn't exist, because left-hand limit != right-hand limit.

    golfinguy:
    Code:
    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.

  9. #24
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    yes, i know tha FToC only applies to continuous functions. I realized it later, thanks.

  10. #25
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    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
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  11. #26
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    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.

  12. #27
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    To be fair to your teacher I think she wanted the problem
    to be
    (sin(x) - .5) / (pi/6 -x) as x->pi/6

  13. #28
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    *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:
    Code:
    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...

  14. #29
    Senior Member joshdick's Avatar
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    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.
    Code:
    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.

  15. #30
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    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).

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