Question

Question #d5319

Answer 1

See other answer.

Answer 2

`2`

Explanation:

`lim_(x->oo) log_e(((3x+1)/(3x-2))^(2x+1)) = log_e(lim_(x->oo)((3x+1)/(3x-2))^(2x+1))`

now making the transformation

`(3x+1)/(3x-2) = 1+1/y` or `x = 1/3(3y+2)` we have

`((3x+1)/(3x-2))^(2x+1) equiv (1+1/y)^(2y+7/3) = (1+1/y)^(7/3)((1+1/y)^y)^2`

now `lim_(x->oo) log_e(((3x+1)/(3x-2))^(2x+1)) = log_e(lim_(y->oo)(1+1/y)^(7/3)lim_(y->oo)((1+1/y)^y)^2) = log_e e^2 = 2`