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Limit at infinity; l'hopital's rule not working as expected

  1. Oct 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Lim(t->(inf)) 1/2((t^2)+1) + (ln|(t^2)+1|)/2 - 1/2

    2. Relevant equations

    N/A (unless L'Hopital's rule can be counted as an equation for this section)

    3. The attempt at a solution

    Background:

    The problem started with:

    inf
    ∫(x^3)/((x^2)+1)^2 dx
    0

    Using partial fraction decomposition, and using two separate online calculators to verify the answer, this came to:


    1/2((x^2)+1) + (ln|(x^2)+1|)/2

    Per requirements for bounds at infinity, I substituted infinity for t, coming out to:


    1/2((t^2)+1) + (ln|(t^2)+1|)/2 - 1/2

    This appears to be something to use L'hopital's rule on. It is in 0 + Infinity state at the beginning. Given the infinity involved, the 1/2 is ignored as it is numerical insignificant as t becomes maximally large.

    Making the appropriate simplification:

    (1 + ((t^2) + 1)(ln|(t^2)+1|))/((2t^2)+2)

    To save a bit of typing, L'hopital's rule, for the first two times, but after t dropped out of the denominator, it was reintroduced in the third derivative set, and seems to go on forever (implying the limit does not exist). When I plug in the original integral with a very large upper bound, it appears to be going to zero. However, I'm not satisfied with that; I need to know what I'm doing wrong in this limit that is causing it to give me an incorrect answer.
     
  2. jcsd
  3. Oct 24, 2012 #2
    What makes you think you can use L'Hopital here? Unless there's a typo in your formula, L'Hopital seems completely inapplicable.

    Perhaps what you need instead is just the sum of two limits: if lim f(x) = a, lim g(x) = b and a+b is well defined (not ∞-∞), then lim (f(x)+g(x)) = a+b.
     
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