# Some asymptotics with log

I would like to answer the question whether $$(1+\frac{\log x}{x})^x\cdot\frac{\log x}{x}-1\sim \log x-1\text{ as }x\to\infty.$$

I think this boils down to the question whether $$(1+\frac{\log x}{x})^x\sim x\text{ as }x\to\infty.$$

I think this is equivalent to $$x\log(1+\frac{\log x}{x})\sim\log x\text{ as }x\to\infty$$

and this should be true since $\log x/x\to 0$ for $x\to\infty$, meaning that $$\log(1+\frac{\log x}{x})\sim\frac{\log x}{x}\text{ as }x\to\infty,$$ hence $$x\cdot\log(1+\frac{\log x}{x})\sim x\cdot \frac{\log x}{x}=\log x\text{ as }x\to\infty$$

• Please, if you are ok, you can accept the answer and set it as solved. Thanks!
– user
Jan 24, 2018 at 22:06

$$\left(1+\frac{\log x}{x}\right)^x=e^{x \log \left(1+\frac{\log x}{x}\right) }\sim e^{\log x}=x$$