How to prove $\lim_{n\to \infty}\frac{n}{\ln{n}}(1-a_{n})=1$? let equation $x^n+x=1$ have positive $a_{n}$.
show that
$$\displaystyle\lim_{n\to \infty}\dfrac{n}{\ln{n}}(1-a_{n})=1$$
yesteday, I have post this and prove following
$$\displaystyle\lim_{n\to \infty}a_{n}=1$$
How prove this limit $\displaystyle\lim_{n\to \infty}a_{n}=1$
Now I found This beautiful  limtit. Thank you everyone can prove it

Let $a_n=1-\varepsilon$.  Then
$$(1-\varepsilon)^n - \varepsilon=0\implies n = \frac{\log{\varepsilon}}{\log{(1-\varepsilon)}} \sim \frac{1}{\varepsilon}\log{\frac{1}{\varepsilon}} \quad (\varepsilon \to 0)$$
One may show that, in this case, because this limit is equivalent to $n \to \infty$,
$$\varepsilon \sim \frac{\log{n}}{n} \quad (n \to \infty)$$
To see this, note that
$$\frac{1}{\varepsilon}\log{\frac{1}{\varepsilon}}\sim \frac{n}{\log{n}} \log{\frac{n}{\log{n}}} = \frac{n}{\log{n}} (\log{n}-\log{\log{n}})\sim n $$
Therefore,
$$\lim_{n \to \infty} \frac{n}{\log{n}} (1-a_n) = \lim_{n \to \infty} \frac{n}{\log{n}} \varepsilon = 1$$
as was to be shown.
