How to prove that if $x_n\to -\infty$ then $\frac{1}{x_n}\to 0$ as $n\to \infty$.
My attempt:
Let $x_n\to -\infty$ and $\epsilon\gt 0$. By the Archimedean Principle pick $N\in \mathbb N$ such that $N\gt \large\frac{1}{\epsilon}$. By taking the reciprocal, we can see that $x_n\ge N$ implies that $\large\frac{1}{x_n}\le \frac{1}{N}\lt \epsilon$. Therefore, since $\large\frac{1}{x_n}$ are all negative, we know that $\left|\large\frac{1}{x_n}\right|\lt \epsilon$ for all $n\ge N$.
This is a similar method I used on a similar problem and am not sure if it is really appropriate or not.
Here is a pedantic but complete proof. Pick any $\varepsilon>0$, and choose $N=N(\varepsilon)\in \mathbb{N}$ such that $x_n<-\frac{1}{\varepsilon}$ whenever $n > N$. This is possible because $x_n \to -\infty$ as $n \to +\infty$. In particular, $x_n<0$ for every $n > N$. For these $n$'s, $$ -\varepsilon<\frac{1}{x_n}<0, $$ and therefore $\lim_{n \to +\infty} \frac{1}{x_n}=0^{-}$.