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Limit question series representation

  1. Jul 28, 2010 #1
    1. The problem statement, all variables and given/known data

    Lim (t->0+) [exp(-x^2/t)] / sqrt(t)

    3. The attempt at a solution

    Ive looked at series representation:

    1/sqrt(t) * [exp(-x^2) - 1+t^2/2 + t^3/6 + ...] but that doesnt help.
    L'Hopital's rule isnt any use either because the t terms are in the denominator but I know this tends to 0
     
  2. jcsd
  3. Jul 28, 2010 #2
    Try making the substitution:
    [tex]t \rightarrow \frac{1}{u}[/tex]
    and taking the limit [tex]u \rightarrow \infty[/tex].

    Retry L'Hopital's rule.
     
  4. Jul 28, 2010 #3

    Mark44

    Staff: Mentor

    Where does the - 1+t^2/2 + t^3/6 + ... part come from?
     
  5. Jul 28, 2010 #4
    hi Mark44, the series comes from splitting out into exp(-x^2) and exp(t) and then using the series representation of the exponential function on the second term.

    The substitution approach works. Thanks Coto
     
  6. Jul 28, 2010 #5

    Mark44

    Staff: Mentor

    But you seem to be saying that e^(a/b) = e^a - e^b. That's not true.
     
  7. Jul 28, 2010 #6
    Sorry you're right. That a mistake
     
  8. Jul 28, 2010 #7
    Coto, actual I don't see how that sub helps because I now have:

    (u^3/2)(x^2/2)exp(-ux^2/4) as u tends to inf

    The exponential term tends to zero however the u^3/2 tends to infinity
     
  9. Jul 28, 2010 #8

    Dick

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    Go back to exp(-x^2/t)*t^(-1/2). Take the log. Now try to figure out what the limit of the log might be. l'Hopital will help with some steps on the way.
     
    Last edited: Jul 28, 2010
  10. Jul 29, 2010 #9
    So if we perform l'hopital first and then log we obtain

    Ln(L)=ln(-x^2)-ln(2t^0.5)+x^2/4t

    The limit of -ln(2t^0.5) is -inf and the limit of x^2/4t is also inf as t -> 0 and exp(inf) = inf which means the limit is indeterminate
     
    Last edited: Jul 29, 2010
  11. Jul 29, 2010 #10
    Is L'hopitals rule the problem in this situation?
     
  12. Jul 29, 2010 #11

    Dick

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    I would do log first. If you get something like (1/t)-log(t) write it as (1-t*log(t))/t and think about that. l'Hopital might help with thinking about the t*log(t) part.
     
  13. Jul 29, 2010 #12
    ln(L) = (-x^2-4tln(sqrt(t))) / 4t

    L'Hopital's: f'(t)/g'(t) where f(t) = -[x^2+4tln(sqrt(t))] and g(t)=4t

    f'(t) = -[4ln(sqrt(t)) + 4t/2x], g'(t) = 4

    f'(t)/g'(t) = -[ln(sqrt(t)) + t/2x] = inf. L = e^inf = inf which is not the correct answer

    Where is this approach going wrong?
     
  14. Jul 29, 2010 #13

    Dick

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    Think about the t*log(sqrt(t)) part separately before you apply l'Hopital to the whole thing. It may NOT even be indeterminant. Write it as log(sqrt(t))/(1/t). Apply l'Hopital to just that. It wouldn't hurt to notice log(sqrt(t))=(1/2)*log(t) either.
     
  15. Jul 29, 2010 #14
    So, just to follow up: log(L) = -[(x^2-4tlog(sqrt(t)))/4t]

    applying l'hopital to 4tlog(sqrt(t)) or 2tlog(t) where f(t) = 2log(t) and g(t)=1/t gives

    -2t^2/t = -2t

    Substituting, our limit becomes log(L) = (-x^2 + 2t)/4t = -x^2/4t + 1/2 = -inf
    exp(-inf) = 0.

    Thank you!
     
  16. Jul 29, 2010 #15

    Dick

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    It might be a little dangerous to be saying things like lim 4tlog(sqrt(t))=(-2t). The limit is a number. Not an expression. I would say the limit is zero and leave it at that.
     
  17. Jul 29, 2010 #16
    Ah yes I see your point. Thanks
     
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