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Limit problem with b^m*e^-(sb)

  1. Sep 14, 2009 #1
    1. The problem statement, all variables and given/known data

    show for m element of Natural numbers lim b->oo b^m/(e^(sb)) = 0



    2. Relevant equations



    3. The attempt at a solution

    Here is what I've started with so far but have ended up at a dead end. I just would like some help to point me in the right direction if possible. Thank you in advance


    as b->oo b^m/(e^(sb)) yields the indeterminate form oo/oo.

    both b^m and e^(sb) are differentiable near the limit

    the derivative of the denominator is non zero so l'hopitals rule is applicable

    However, when I apply lhopital's rule I get

    lim b->oo mb^(m-1)/(se^(sb))

    My approach was to prove this inductively, ie for m=1 and go from there but I can't even show this is true for m=1. Maybe this is completely the wrong approach. I have convinced myself that I am heading in the wrong direction.

    Thank you.
     
  2. jcsd
  3. Sep 14, 2009 #2

    lanedance

    User Avatar
    Homework Helper

    i think you're heading in the right dierection

    after using L'hop's rule & differntiating you have a similar limit except with now ith b^m-1, ie. one lower power of b on numerator

    what happens when you have reapeat L'hop's rule m times?
     
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