calc
posted by lola .
Use L’Hopital’s rule to find the limit of this sequence
(n^100)/(e^n)
...If you do L'Hop. Rule it would take forever, right? You would always get an (e^n) at the bottom and will have to use the L'Hop. rule 100 times to find the limit...100*n^99, 9900n^98, and etc.
Is there a shortcut to find the limit?
Or, am I doing something way wrong?
calc  Reiny, Wednesday, February 1, 2012 at 12:06pm
You didn't say what the aprroach value is for n
is n> 0 ?
lim n^100/e^n)
= lim (100n^99)/(e^n)
= lim (9900n^98) / e^n
= ...
= lim (huge n^1)/e^n
= lim (more huge )/e^n
= lim (0/e^n)
= 0/1
= 0
calc  lola, Wednesday, February 1, 2012 at 12:29pm
what if n approaches infinity?

This shows that any polynomial, no matter how high the degree, grows more slowly than any exponential function of a base >1.
Repeated application of L’Hopital’s Rule just reduces the degree of the numerator by 1, and when you get to a constant, one more application reduces the numerator to ZERO. The denominator is still e^x. 
so the limit is zero

is the limit zero?
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