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A seemingly tricky limit - L'Hopital

  1. May 19, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex] \lim_{t \to \infty} \frac {1-\frac{t}{(t-1)}}{1-\sqrt{\frac{t}{(t-1)}}} [/tex]

    2. Relevant equations



    3. The attempt at a solution
    I am pretty sure that everything I did here was legal, I just wanted to check. I got the right answer, so yeah, heres what I did:

    [tex] \lim_{t \to \infty} \frac {1-\frac{t}{(t-1)}}{1-\sqrt{\frac{t}{(t-1)}}} [/tex]

    derivative turns out to be:

    [tex] \lim_{t \to \infty} \frac{\frac{1}{(t-1)^{2}}}{\frac{1}{2}(\frac{1}{(t-1)^{2}})(\frac{t}{t-1})^{-1/2}} [/tex]

    and after the obvious term cancels we have:

    [tex] \lim_{t \to \infty} \frac {1}{\frac{1}{2} (\frac{t}{t-1})^{-1/2}} [/tex]

    and since it is a reciprocal of a reciprocal this is the same thing:

    [tex] \lim_{t \to \infty} 2(\frac{t}{t-1})^{1/2} [/tex]

    then with the limit applied:


    [tex] \lim_{t \to \infty} 2(\frac{t}{t-1})^{1/2} = 2[/tex]

    the step with the flipping of the reciprocals is really all I need check, I am fairly certain everything else is good.

    thanks!
     
  2. jcsd
  3. May 19, 2010 #2

    Dick

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    Looks fine to me. Well done.
     
  4. May 19, 2010 #3
    Make a substitution:

    [tex]

    u = \sqrt{\frac{t}{t - 1}}

    [/tex]

    and re-express your expression in terms of [itex]u[/itex]. When [itex]t \rightarrow \infty[/itex], what does [itex]u[/itex] tend to? You will have a simpler problem that can be solved by elementary methods (factorization).
     
  5. May 19, 2010 #4
    I used a difference of squares and Dickfore's substitution to make it (1+u)(1-u) over (1-u). Why did you take the derivative? It just looks confusing?
     
  6. May 19, 2010 #5

    Dick

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    Asphyxiated solved it by using l'Hopital on a 0/0 form. The given solution is just fine. It's simpler with a substitution, but that doesn't mean there is anything wrong with what Asphyxiated did.
     
  7. May 19, 2010 #6
    Oh! Sorry! I meant no harm, and I certainly didn't mean to imply that there was anything wrong with Asphyxiated's solution. My post may have sounded rude but really I was just surprised. I had temporarily forgotten about l'Hopital's rule because we aren't allowed to use it in my grade 12 class, and when I saw this limit I just thought of difference of squares right away.
     
  8. May 19, 2010 #7
    I did it the way that I did it because it is in the section of the book to learn to use l'hopital's rule, that is it.
     
  9. May 19, 2010 #8

    Dick

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    I didn't think you were trying to be rude or anything, don't apologize. But you asked "why the derivative", so I answered.
     
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