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

1. Sep 14, 2009

fleuryf

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. Sep 14, 2009

lanedance

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?