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?

If n approaches infinity, we can still use L'Hopital's rule to find the limit of the sequence (n^100)/(e^n). Here's how:

1. Take the derivative of the numerator and the denominator:
- The derivative of n^100 is 100n^99.
- The derivative of e^n is e^n.

2. Take the limit as n approaches infinity of the derivative of the numerator divided by the derivative of the denominator:
- lim (n^100)/(e^n) = lim (100n^99)/(e^n)
- As n approaches infinity, both the numerator and denominator grow exponentially, so we can use L'Hopital's rule repeatedly.

3. Apply L'Hopital's rule until we reach an expression that can be directly evaluated:
- lim (100n^99)/(e^n) = lim (9900n^98)/(e^n)
- lim (9900n^98)/(e^n) = lim (980100n^97)/(e^n)
- ...
- lim (0)/(e^n) = 0

Therefore, the limit of the sequence (n^100)/(e^n) as n approaches infinity is 0.