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; Excuse me for bumping... I'm just adding new info to the thread. My teacher gave me a name of a ...

  1. #31
    Just because ygfperson's Avatar
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    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?

  2. #32
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    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

  3. #33
    Just because ygfperson's Avatar
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    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

  4. #34
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    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.
    Last edited by alpha; 02-12-2003 at 03:46 PM.

  5. #35
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    >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....?
    Last edited by *ClownPimp*; 02-12-2003 at 05:04 PM.
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  6. #36
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    Also,

    >5) Assume that f(a) = g(a) = 0
    >lim b->a f'(c)/g'(c) == (f(b)/g(b))

    Code:
    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
    C Code. C Code Run. Run Code Run... Please!

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  7. #37
    Just because ygfperson's Avatar
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    d'oh... #5 was a typo. Or I just forgot to include the limit.

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