Posted by Help!! on Saturday, December 1, 2012 at 9:27pm.
take the limit as x goes to infinity: (14x/(14x+3))^(4x)

calculus  Damon, Saturday, December 1, 2012 at 9:54pm
x = 1
f = (14/17)^4 = .46
x = 10
f = (140/143)^40 = .428
x = 100
f = (1400/1403)^400 = .4248
x = 1000
f = (14,000/14,003)^4000 = .4244
hmmm 
calculus  Damon, Saturday, December 1, 2012 at 10:01pm
x = 10,000
f = (140,000/140,003)^40,000 = .4244
double hmmm 
calculus  Damon, Saturday, December 1, 2012 at 10:03pm
x = 100,000
f = (1,400,000/1,400,003)^400,000 = .4244 
calculus  Count Iblis, Saturday, December 1, 2012 at 10:45pm
Log[f(x)] =
4 x [log(14 x)  log(14 x + 3)] =
4 x log[1 + 3/(14 x)] =
4 x [3/(14 x) + O(1/x^2)] =
6/7 + O(1/x)
The limit for x to infinity of
log[f(x)] is thus 6/7, the limit of
f(x) is thus exp(6/7) 
calculus  Help!!, Sunday, December 2, 2012 at 2:58pm
What happened between
4 x log[1 + 3/(14 x)] =
and
4 x [3/(14 x) + O(1/x^2)] =
How did you get rid of the log? Thx in advance