$[1-x_n]^{n}\to 1$ implies that $n x_n\to 0$ as $n\to\infty$? As the title says $[1-x_n]^{n}\to 1$ implies that $n x_n\to 0$ as $n\to\infty$?
We know that $\left(1+\frac{x}{n} \right)^n \to e^x$ as $n\to\infty$. This implies (not sure why) that  $\left(1+\frac{x}{n} +o(1/n) \right)^n \to e^x$.
So the result obtains since we must have that $x_n=\frac{x}{n}+o(1/n)$ where $x=0$?

For $n$ be sufficiently large, we have $(1-x_n)^n >0$ (since it converges to 1).
We can then say that $n \log(1-x_n)$ tends to zero. Necessarily $\log(1-x_n)$ tends to zero.
So $x_n$ tends to zero and we have that :   $ \quad \log(1+x) \underset{0}{\sim} x$
Then $nx_n \to 0$.
