TITLE: Prove $\lim\limits_{n\to\infty}(\frac{a}{n})=0$
QUESTION [6 upvotes]: I'm very new to the epsilon definition of a limit and I was hoping I could get some feedback to see if I have the right idea.
For any $a\in\mathbb{R}$, $$\lim\limits_{n\to\infty}\left(\frac{a}{n}\right)=0$$
Proof:
Let $\epsilon >0$. Choose $N>\frac{|a|}{\epsilon}$. Then for all $n\geq N$, since $n\geq N>\frac{|a|}{\epsilon}$, we have
$$n>\frac{|a|}{\epsilon}\;\;\;\Rightarrow\;\;\;\epsilon >\frac{|a|}{n}$$
Therefore,
$$\left|\frac{a}{n}-0\right|=\frac{|a|}{n}<\epsilon$$
$\blacksquare$

REPLY [3 votes]: Your answer is $100\%$ correct. One can generalize your theorem for a sequence $\displaystyle\sigma_{n}=\frac{\alpha}{n^{r}}$ with $\alpha\in\mathbb{R}$ and $r\in\mathbb{R}^{+}$ as follow :
Let $\varepsilon>0$, then by the archimedean property, $\exists N\in\mathbb{N}$ such that $\displaystyle\frac{\alpha}{N}<\sqrt[r]{\varepsilon}$. Now we show that starting this index $N$, elements of $\sigma_{n}$ are in the neighborhood of $0$. For $n\geq N$, we have that :
$$
\left|\frac{\alpha}{n^{r}}\right|=\frac{|\alpha|}{n^{r}}\leq\frac{|\alpha|}{N^{r}}<\varepsilon
$$
which means that $|\sigma_{n}|<\varepsilon\implies\displaystyle\lim_{n\to\infty}\sigma_{n}=0$