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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 $$ x\cdot\log(1+\frac{\log x}{x})\sim x\cdot \frac{\log x}{x}=\log x\text{ as }x\to\infty $$

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

1 Answer 1

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Note that

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

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