We have lim_(xrarroo)(ln(2x)-ln(1+x))
Since lna-lnb=ln(a/b), we write:
lim_(xrarroo)ln((2x)/(1+x))
According to the limit chain rule, let's say we have a function (f@g), or f(g(x)). The limit of this function, when x tends to a, is given by L. To find L, we first find lim_(xrarra)g(x). The answer to this is given as b. Then, we find lim_(urarrb)f(u), where u=g(x). The answer for this final limit is L.
Here we take f(x)=lnx and g(x)=(2x)/(1+x). Also, here, a=oo. First we take:
lim_(xrarroo)(2x)/(1+x)
2lim_(xrarroo)x/(1+x)
We input oo:
2(oo/oo)
We see that we get an indeterminate form. According to L'Hopital's Rule, when lim_(xrarrc)f(x)/g(x) yields an indeterminate form, such as oo/oo, the limit is given by lim_(xrarrc)(f'(x))/(g'(x)).
We go back to lim_(xrarroo)x/(1+x). Since d/dxx=1 and d/dx(1+x)=1, we write it as:
2*(1/1)
2(1)
=2
So we say that b=2. We compute:
lim_(urarr2)ln(u)
Simply input and we get:
ln(2)
Our answer.
