First, we use the property of logarithms that #log_a(x)-log_a(y) = log_a(x/y)#
#lim_(x->oo)ln(x^2-1)-ln(2x^2+3) = lim_(x->oo)ln((x^2-1)/(2x^2-3))#
Next, we use the property of continuous functions that if #f(x)# is continuous on an interval containing #a# and #lim_(x->a)g(x)# exists, then #lim_(x->a)f(g(x)) = f(lim_(x->a)g(x))#. As the natural log #ln(x)# is continuous on #(0, oo)#, that means
#lim_(x->oo)ln((x^2-1)/(2x^2-3)) = ln(lim_(x->oo)(x^2-1)/(2x^2-3))#
Now we just need to evaluate the limit of the rational expression within the logarithm.
#lim_(x->oo)(x^2-1)/(2x^2+3) = lim_(x->oo)(1-1/x^2)/(2+3/x^2)#
#=(1-1/oo)/(2+3/oo)#
#=(1-0)/(2+0)#
#=1/2#
Substituting that back into the logarithm, we get our final result:
#lim_(x->oo)ln(x^2-1)-ln(2x^2+3)=ln(lim_(x->oo)(x^2-1)/(2x^2-3))#
#=ln(1/2)#
(We could also use the property that #log(a^x) = xlog(a)# to express the answer as #ln(1/2) = ln(2^(-1)) = -ln(2)#)
