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Homework Help: L'Hospitals problem NEED HELP

  1. Dec 4, 2008 #1
    L'Hospitals problem...NEED HELP!!!

    This problem is on a test that our teacher said we could research if we wanted...and we are finishing it tomorrow. I have NO CLUE how to approach it and need help!

    1. The problem statement, all variables and given/known data
    lim x->infinity [tex]
    x - ln(1+2e^x)
    [/tex]

    2. Relevant equations



    3. The attempt at a solution

    so, its infinity - infinity, which is indeterminate.

    I did the limit function on my TI-89, and it just keeps spitting the original equation back at me as the answer. I then graphed the function on my TI-89, and when it goes past x=2302, the graph is undefined. HOWEVER, when i go into the table, the values after 2302 go to -infinity, so I think thats what the answer's supposed to be...but I have no idea how to set up the quotient and solve using L'Hospital's Rule...PLEASE HELP!!!
     
  2. jcsd
  3. Dec 4, 2008 #2

    Dick

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    Re: L'Hospitals problem...NEED HELP!!!

    Don't use l'Hopital on that! Write (1+2e^x)=(e^x)*(2+1/e^x) and use rules of logs.
     
  4. Dec 4, 2008 #3
    Re: L'Hospitals problem...NEED HELP!!!

    ...i dont quite follow where you're going with that...can you elaborate more please?
     
  5. Dec 4, 2008 #4

    Dick

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    Re: L'Hospitals problem...NEED HELP!!!

    It's your turn to elaborate. Simplify ln((e^x)*(2+1/e^x)).
     
  6. Dec 4, 2008 #5
    Re: L'Hospitals problem...NEED HELP!!!

    [tex]
    x - {ln[(e^x)(2 + 1/e^x)]}
    [/tex]

    rule of logs:
    [tex]
    x - [ln(e^x) + ln(2 + 1/e^x)]
    [/tex]

    cancel out ln(e^x)
    [tex]
    x - x - ln(2 + 1/e^x)
    [/tex]

    plug in the limit:

    [tex]
    =infinity- 1 - ln(2)
    [/tex]

    [tex]
    =infinity
    [/tex]

    +infinity is the answer? when my calc says -infinity?
     
    Last edited: Dec 4, 2008
  7. Dec 4, 2008 #6
    Re: L'Hospitals problem...NEED HELP!!!

    ln(e^x)=x; not 1.

    final solution shoud be ln(2)
     
  8. Dec 4, 2008 #7
    Re: L'Hospitals problem...NEED HELP!!!

    Oops...dumb mistake. My bad. I feel like a retard. :(

    ohhh...ok I think I got it. Thanks to all! :)
     
    Last edited: Dec 4, 2008
  9. Dec 4, 2008 #8

    jambaugh

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    Re: L'Hospitals problem...NEED HELP!!!

    To apply l'Hospital's rule recall that since [tex]e^x[/tex] is a continuous function you can "move" limits in and out of an exponential.

    Thus the limit you want being [tex]L=\lim_{x\to \infty} f(x) -g(x)[/tex],

    [tex]\exp({\lim_{x\to \infty} f(x) -g(x))= \lim_{x\to \infty} e^{f(x)-g(x)} [/tex]

    Then apply rules of exponentials:
    [tex]\exp(\lim_{x\to \infty} f(x) -g(x))= \lim_{x\to \infty} \frac{e^{f(x)}}{e^{g(x)}}[/tex]

    You can now apply l'Hospital's rule on this limit of a quotient and the answer is the exponential of your desired limit. (if it is finite and positive).

    Now there may be a more direct way to calculate but this is how you deal with differences of infinities in general so you will want to practice this method.

    There is a bit more to it... I should rather have written, given the exponential function is continuous:
    [tex] \lim_{x\to a}e^{h(x)} = \lim_{y\to L} e^y[/tex] where [tex] L = \lim_{x\to a} h(x)[/tex].
    But this is the same thing provided the limit L is finite. It just generalizes to the case where L is infinite.

    But the result is that for your limit:
    [tex] \lim_{x\to \infty} f(x) -g(x) = \ln\left[ \lim_{x\to \infty} \frac{e^{f(x)}}{e^{g(x)}}\right] [/tex]
    provided this logarithm is well defined.
     
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