1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    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


    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


    User Avatar
    Science Advisor
    Homework Helper

    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


    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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Limit question series representation
  1. Series representation (Replies: 3)