$[1-x_n]^{n}\to 1$ implies that $n x_n\to 0$ as $n\to\infty$?

Question

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 $x=0$?

Answer 1

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 $nx_n \to 0$.

Answer 2

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 $-\epsilon<nx_{n}<\epsilon$.