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Math Help - l'hospitals rule help

  1. #1
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    l'hospitals rule help

    i'm having a lot of trouble with these two problems.

    the first one is tough, and i don't know how to do it, i just know the answer is supposed to be 1/2:

    lim ((1/ln x) - (1/(1-x))
    x --> 1

    i found the derivatives of both and got:

    ((-1/(x ln x ^2) - ((1)/(x-1)^2)

    but i don't know if that's what i'm supposed to do

    and for the second one, i don't even know where to start:

    lim (1 + (a/x)) ^ bx
    x --> infinity

    please....help. these are the last two of my homework problems and they're killing me!
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  2. #2
    Junior Member rubix's Avatar
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    remember, to apply L'Hospital's Rule you need

    lim x --> c  f(x)/ g(x) where f(x) and g(x) are some function dependent on x. Furthermore, lim f(x) and lim g(x) both must be either 0 or +- infinity. This mear fact tell you that, you want to have a fraction of two function with x on numerator and denominator such that both have limit or either 0 or +- infinity.

    once that is done, following holds:

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


    so what i'm trying to tell you here is this:

    a) you want to have numerator and a denominator dependence on x. So, i would start by taking common denominator to get something of the form f(x) / g(x)

    a) you want to make sure both lim f(x) and lim g(x) go to 0 or +- infinity before you use L'Hospital's Rule. So, taking derivative as you did is irrelevant when you clearly don't have a single numerator and a denominator and neither they go to 0 or +- infinity in given form.
    Last edited by rubix; November 15th 2009 at 03:24 PM.
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  3. #3
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    Quote Originally Posted by boomersooner1331 View Post
    i'm having a lot of trouble with these two problems.

    the first one is tough, and i don't know how to do it, i just know the answer is supposed to be 1/2:

    lim ((1/ln x) - (1/(1-x))
    x --> 1

    i found the derivatives of both and got:

    ((-1/(x ln x ^2) - ((1)/(x-1)^2)
    Well I know that you can only apply l'hopital's rule to an undeterminate form (0/0) or (infinity / infinity). I'm currently working on trying to solve it =D
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  4. #4
    Junior Member rubix's Avatar
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    i had enough time in my hand....

    Quote Originally Posted by boomersooner1331 View Post

    lim ((1/ln x) - (1/(1-x))
    x --> 1
     1/lnx - 1/ (1-x) = (1-x - lnx) / lnx (1-x)

    notice lim x --> 1 of top is 0 and bottom is 0 as well. Use L'Hospitals.

    Quote Originally Posted by boomersooner1331 View Post
    lim (1 + (a/x)) ^ bx
    x --> infinity
    [math(1 + a/x) ^ (bx) = (x +a) ^ (bx) / (x)^bx

    notice as x -> inf top and bottom goes to infinity. You know what to do next

    P.S. i have not worked out the solution so idk how it works then forth...you might have to use L'hospitals rule once more.
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  5. #5
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    Quote Originally Posted by rubix View Post
    i had enough time in my hand....



     1/lnx - 1/ (1-x) = (1-x - lnx) / lnx (1-x)

    notice lim x --> 1 of top is 0 and bottom is 0 as well. Use L'Hospitals.
    But if you go from common factor and use L'Hopital's rule you still get something that's not easily resolvable....
    Last edited by Veraltus; November 15th 2009 at 03:35 PM. Reason: misformatted
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  6. #6
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    the second one is just one since they have the same powers and the coefficients are 1, right?

    sorry if that seems like a dumb question. i'm absolutely exhausted. this has been the longest weekend of my life
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  7. #7
    Junior Member rubix's Avatar
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    Quote Originally Posted by Veraltus View Post
    But if you go from common factor and use L'Hopital's rule you still get something that's not easily resolvable....
    then use probably have to use L'Hospital's rule once more. In that case, you'll prolly need to use product rule on bottom.
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  8. #8
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by boomersooner1331 View Post
    and for the second one, i don't even know where to start:

    lim (1 + (a/x)) ^ bx
    x --> infinity

    please....help. these are the last two of my homework problems and they're killing me!
    (1+a/x)^{bx}=\exp\left[bx\ln(1+a/x)\right]

    Now you need to find the limit of the stuff inside the \exp.

    \lim_{x\to\infty}\frac{b\ln(1+a/x)}{x^{-1}}=\lim_{x\to\infty}\frac{\frac{-ab}{x(x+a)}}{-x^{-2}}= \lim_{x\to\infty}\frac{abx^2}{x(x+a)}=\lim_{x\to\i  nfty}\frac{ab}{1+a/x^2}=ab

    So \lim_{x\to\infty}(1+a/x)^{bx}=\lim_{x\to\infty}\exp\left[bx\ln(1+a/x)\right]= \exp\left[\lim_{x\to\infty}bx\ln(1+a/x)\right]=\exp(ab)=\boxed{e^{ab}}
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  9. #9
    Super Member redsoxfan325's Avatar
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    Actually, I just thought of an easier way to do this:

    \lim_{x\to\infty}\left(1+\frac{a}{x}\right)^{bx}=\  lim_{x\to\infty}\left[\left(1+\frac{1}{x/a}\right)^{x/a}\right]^{ab}=\left[\lim_{x\to\infty}\left(1+\frac{1}{x/a}\right)^{x/a}\right]^{ab}

    We know that e is defined to be the stuff inside the brackets, so the final answer is e^{ab}.
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