Prove that $\Bigl( 1 - \frac{1}{x} \Bigr)^{\frac{x}{\ln x}} = 1 + O \Bigl( \frac{1}{\ln x} \Bigr)$

Question

It was my first attempt to use $\lim\limits_{x \rightarrow \infty} \Bigl( 1 - \frac{1}{x} \Bigr)^{x} = e^{-x}$. But then I didn't know what to do with $\frac{1}{\ln x}$. On my second try, I wanted to use \begin{align*} \Bigl( 1 - \frac{1}{x} \Bigr)^{\frac{x}{\ln x}} - 1 = e^{-x} \Bigl( e^{x \frac{\ln(x-1)}{\ln(x)}} - e^x \Biggr). \end{align*} But the attempt was unsuccessful, because I didn't know how to estimate the term to prove the statement.

Answer 1

It suffices to prove that $$1 - \left(1 - \frac{1}{x}\right)^{\frac{x}{\ln x}} \le \frac{1}{\ln x},\ \forall x \ge \mathrm{e}^2.$$ It suffices to prove that $$\left(1 - \frac{1}{\ln x}\right)^{\ln x} \le \left(1 - \frac{1}{x}\right)^{x}, \ \forall x \ge \mathrm{e}^2.$$

Let $f(u) = (1 - \frac{1}{u})^u$. We have $$f'(u) = (1 - \frac{1}{u})^u \left(\ln\left(1 - \frac{1}{u}\right) + \frac{1}{u-1}\right) > 0, \ \forall u\ge 2.$$

Since $2 \le \ln x < x$ for $x\ge \mathrm{e}^2$, we have $f(\ln x) < f(x)$.

We are done.

Answer 2

Hint

Try to use

$$y=\Big[ 1 - \frac{1}{x} \Big]^{\frac{x}{\log( x)}}\implies \log(y)= {\frac{x}{\log(x)}}\log\Big( 1 - \frac{1}{x} \Big)\sim -\frac{1}{\log(x)}$$ $$y=e^{\log(y)}\implies y\sim\cdots ???$$