Question Number 125523 by Algoritm last updated on 11/Dec/20
Answered by Dwaipayan Shikari last updated on 11/Dec/20
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{2}{x}+\mathrm{3}\right){log}\left(\frac{{x}+\mathrm{2}}{{x}}\right) \\ $$$$=\left(\mathrm{2}{x}+\mathrm{3}\right)\left(\frac{\mathrm{2}}{{x}}\right)=\mathrm{4}+\frac{\mathrm{6}}{{x}}=\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{log}\left(\mathrm{1}+{z}\right)={z} \\ $$