Question

computing the limit of 2^n/n^n as n approaches infinity?

Answer 1

using ln(2^n/n^n) results in indeteriminate.

Answer 2

have you seen l'hopital's rule, or no?

Answer 3

It approaches 0, as the denominator approaches infinity faster than the numerator.

Answer 4

Yeah, but that doesnt show any work really.

Answer 5

Oh damnit.

Answer 6

Yes it does. It is called comparing exponential growth! 2^n approaches infinity slower than n^n.

Answer 7

well you can assume that because any number over infinity approaches 0

Answer 8

but we're required to find it either by lhospital or using the ln

Answer 9

skay..no such thins as over infinity.

Answer 10

thing*

Answer 11

1/infinity is 0

Answer 12

thats what i was referring to

Answer 13

This problem is not called 1/infinity...It is called the denominator approaching infinity at a faster RATE than the numerator. They both approach infinity, but the denominator approaches infinity at a faster rate.

Answer 14

okay so with l hospital dont i end up with 2^n(ln(2)/n^n(ln(n)+1)

Answer 15

\[\ln(\frac{ 2^{n} }{ n^{n} }) = \ln[(\frac{ 2 }{ n })^{n}] = nln(\frac{ 2 }{ n })\]

Answer 16

\[lny = \frac{ \ln(\frac{ 2 }{ n }) }{ \frac{ 1 }{ n } }\]derivaitve of ln y is just 1/y of course. the numerator becomes:
\[\frac{ \frac{ -2 }{ n^{2} } }{ \frac{ 2 }{ n } }= \frac{ -1 }{ n }\]derivative of 1/n on bottom becomes -1/n^2. Now simplifying the derivative of the numerator divided by the derivative of the denominator:
\[\frac{ \frac{ -1 }{ n } }{ \frac{ -1 }{ n^{2} } }= n\]So now you just have
\[\frac{ 1 }{ y }= n \implies y = \frac{ 1 }{ n }\]So doing l'hopital's rule gets you here, where the limit is now even more clear than before. I left out the whole limit portion on the left just to save time, but same thing.

Answer 17

thank you!

Answer 18

yep, np :3