lim_(x->0)((3+x)/(3-2x))^(1/x)=lim_(x->0)((3-2x+2x+x)/(3-2x))^(1/x)=
=lim_(x->0)(1+(3x)/(3-2x))^(1/x)=A
Let (3x)/(3-2x)=1/t, then:
3tx=3-2x => x(3t+2)=3 => 1/x=(3t+2)/3=t+2/3
It's obvious that when x->0 then t->oo.
A=lim_(t->oo)(1+1/t)^(t+2/3)=
=lim_(t->oo)(1+1/t)^(2/3) * lim_(t->oo)(1+1/t)^t=
=1 * e=e
Note:
t->oo => 1/t->0
lim_(t->oo)(1+1/t)^t =e
Note 2:
lim_(x->0)((3-2x+2x+x)/(3-2x))^(1/x)=
lim_(x->0)((3-2x)/(3-2x)+(2x+x)/(3-2x))^(1/x)=
lim_(x->0)(1+(3x)/(3-2x))^(1/x)