Does there exist such sequence? Does there exist $\{x_n\}\subseteq (0,1)$ such that \begin{gather*} x_n\rightarrow 1,\\ x_n^n\rightarrow 0\end{gather*} and $$n-\frac{1-x^n_n}{1-x_n}\rightarrow 0?$$
Thanks.
 A: It is not possible. Write $y_n=1-x_n$; what you are looking for is a sequence $y\in(0,1)^\mathbb{N}$ such that


*

*$y_n \xrightarrow[n\to\infty]{} 0$

*$(1-y_n)^n\xrightarrow[n\to\infty]{} 0$

*$\frac{1-(1-y_n)^n}{y_n} = n+o(1)$


Rewriting the last condition, you get $$ ny_n = 1- (1-y_n)^n + o(y_n)  $$
and since the first two imply $(1-y_n)^n=o(1)$ and $y_n = o(1)$, this in turn requires $$ny_n = 1 + o(1).$$ That is, $y_n = \frac{1}{n} + o\left(\frac{1}{n}\right)$. But then, you cannot get the second condition, as you now have $$(1-y_n)^n\xrightarrow[n\to\infty]{}\frac{1}{e}.$$
A: This is impossible: Let's suppose $x_n \to 1$ and $x_n^n \to 0$ , we will see that these two conditions lead to
 $$\displaystyle\liminf_{n\to +\infty}\left( n - \dfrac{1-x_n^n}{1-x_n} \right) \geq1$$
Proof:
Firstly, by $x_n^n \to 0$ we get $x_n < 1$ when $n$ is greater then a certain integer $N$.
Then we have for all $n > N$
\begin{align}
n - \dfrac{1-x_n^n}{1-x_n} &= n - \dfrac{(1-x_n)(\sum_{k=0}^{n-1}x_n^k)}{1-x_n} \\
&=n -\sum_{k=0}^{n-1}x_n^k \\
&= \sum_{k=0}^{n-1}(1-x_n^k)  \geq 1 - x_n^{n-1}
\end{align}
since all $x_n^k<1$
Besides, we have $$x_n^{n-1} = \dfrac{x_n^n}{x_n} \to \frac{0}{1}=0$$ 
Finally we get $$\displaystyle\liminf_{n\to +\infty}\left( n - \dfrac{1-x_n^n}{1-x_n} \right)\geq \liminf_{n\to +\infty} \left( 1 - x_n^{n-1}\right)=1$$
Added:
I'd like to mention by the similar idea in the proof, we can actually prove $$\displaystyle\lim_{n\to +\infty}\left( n - \dfrac{1-x_n^n}{1-x_n} \right) =+\infty$$
