matematykaszkolna.pl
granica asss: [(n−6)/(n+4)]2n
25 mar 22:05
Janek191:
  n − 6  1 −6n 
an = [
]2n = [
]2n =
  n + 4 1 + 4n 
  [ 1 − 6n]n]2 
=
 [1 + 4n]n]2 
więc
  [ e−6]2 
lim an =
= [ e−10]2 = e−20
  [ e4]2 
n→∞
26 mar 00:08