what's the limit of this sequence ?

Question

hey guys : a problem with this sequence : u(n) = n²(2/3)^n
i found in a book that :
lim n²(2/3)^n = o
but i don't get it, cause as far as i know :
lim n² = +infinity and lim (2/3)^n = 0 lim n²(2/3) = +infinity x 0 wich is an indefinite form....
help me know why it would be 0, i'd really appreciate it ...

Answer 1

This is the indeterminate form 0*infinity.
Write it as n^2/1/(2/3^n) so it's form is now infinity/infinity and L'Hospital's rule can be used
lim = 2n/1/-(2/3^n * ln(2/3)/2/3^2n
= lim 2/ (expression that goes to infinity)
so lim = 0

Answer 2

The square term, n^2 tends to infinity slower than the
exponential term, (2/3)^n tends towards zero.

Exponential growth or decay are among the fastest kinds.

Answer 3

Which is the strongest ? n² or (2/3)^n ?
take the log :
log (n² (2/3)^n) = log (n²) + log((2/3)^n)
=2 log (n) + n log (2/3)
As n is strongest than log (n) then
lim log (n² (2/3)^n) = - infinity
and lim (n² (2/3)^n) = 0

Answer 4

well lets take the definition of a limit tending to a number. i take it in this case its infinity.

why dont u substitute a realllly huge number in for n( say 999999999) on your calculator to confirm / change your answer
hope that help s you come to a decision.