Question

take the limit as x goes to infinity: (14x/(14x+3))^(4x)

Answer 1

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

Answer 2

x = 10,000
f = (140,000/140,003)^40,000 = .4244
double hmmm

Answer 3

x = 100,000
f = (1,400,000/1,400,003)^400,000 = .4244

Answer 4

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)

Answer 5

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