First, keep in mind the properties of logarithms.

So, first simplify the function.

Assume f(x)=ln(2x)−ln(1+x)�(�)=ln(2�)-ln(1+�).

Using properties of natural logarithms, the function becomes

f(x)=ln(2x1+x)�(�)=ln(2�1+�)

Then,

f(x)=ln(2)+ln(x1+x)�(�)=ln(2)+ln(�1+�).

Finally, when we take the limit as x approaches infinity, we can take the sum of the limits.

limx→∞f(x)=limx→∞[ln(2)]+limx→∞[ln(x1+x)]lim�→∞�(�)=lim�→∞[ln(2)]+lim�→∞[ln(�1+�)]

limx→∞[ln(2)]=ln(2)lim�→∞[ln(2)]=ln(2)

limx→∞[ln(x1+x)]=ln(limx→∞[x1+x])=ln(1)=0lim�→∞[ln(�1+�)]=ln(lim�→∞[�1+�])=ln(1)=0

ln(2)+0=ln(2)ln(2)+0=ln(2)

Therefore, the limit as x approaches infinity is ln(2).