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))
