so the base is 1+u(x) and u(x)⟶0 as x⟶∞. For small u,ln(1+u)=u+O(u2). Thus lnL=(3x−2)ln(1+u(x))≈(3x−2)u(x). Compute the limit of (3x−2)u(x) : (3x−2)u(x)=(3x−2)⋅
−3x+5
3x2+x−2
=
(3x−2)(−3x+5)
3x2+x−2
. Expand numerator: (3x−2)(−3x+5)=−9x2+21x−10 So (3x−2)u(x)=
−9x2+21x−10
3x2+x−2
∼
−9x2
3x2
=−3 as x⟶∞. More precisely, the limit is
lim
x⟶∞
(3x−2)u(x)=−3 The extra O(u2) term is of order (3x−2)u(x)2∼ const ⋅x⋅
