Question

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

Answer 1

See below.

Explanation:

`log_e(3x+1)-log_e(2x+1)=log_e((3x+1)/(2x+1))` then

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