$[1-x_n]^{n}\to 1$ 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 $\epsilon>0$ we have $\left(1-\frac{\epsilon}{n}\right)^{n}\rightarrow e^{-\epsilon}<1$
and $\left(1+\frac{\epsilon}{n}\right)^{n}\rightarrow e^{\epsilon}>1$.
So $\left(1-\frac{nx_{n}}{n}\right)^{n}=\left(1-x_{n}\right)^{n}\rightarrow1$
tells us that eventually $-\epsilon<nx_{n}<\epsilon$.
