Find the limit as x->infinity of ((3x+2)/(3x+4))^(3x+1)

Answer 1

multiply numerator and denominator by

(1/x)^(3x+1)

Lim ((3+2/x)/(3+4/x))^(3x+1)

so as x>>inf this reduceds to
((3/3)^3x+1) and as x>>inf
= 1

Answer 2

Thank You. This works,and easier I also did it by finding the ln of the expression
(3x+1)[ln(3x-2)-ln(3x+4)]

changed to [ln(3x-2)-ln(3x+4)]/(1/(3x+1)
then used L'Hospital Rule to get lim=0

so since ln y = 0, the e^0=1