Question Number 150759 by nimnim last updated on 15/Aug/21
$$\:\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{log}\left({e}+{x}\right)−\mathrm{1}}{{x}}=? \\ $$$${please}\:{help}.. \\ $$
Answered by Olaf_Thorendsen last updated on 15/Aug/21
$$\frac{\mathrm{ln}\left({e}+{x}\right)−\mathrm{1}}{{x}}\:=\:\frac{\mathrm{ln}{e}+\mathrm{ln}\left(\mathrm{1}+\frac{{x}}{{e}}\right)−\mathrm{1}}{{x}} \\ $$$$=\:\frac{\mathrm{ln}\left(\mathrm{1}+\frac{{x}}{{e}}\right)}{{x}}\:\underset{\mathrm{0}} {\sim}\:\frac{\frac{{x}}{{e}}}{{x}}\:=\:\frac{\mathrm{1}}{{e}} \\ $$
Commented by nimnim last updated on 15/Aug/21
$${Thank}\:{you}\:{so}\:{much}.,{Sir}. \\ $$