TITLE: My proof of $\lim_{n\to\infty} (1 / n) = 0$
QUESTION [7 upvotes]: Is my proof correct?
We consider the sequence
  \begin{equation*}
    (x_n)_{n = 1}^{\infty},
    \qquad \text{where} \qquad
    x_n = \frac{1}{n}.
  \end{equation*}
$\textbf{Theorem.}$
  \begin{equation*}
    \lim_{n \to \infty} x_n = 0.
  \end{equation*}
$\textit{Proof.}$
  By definition, we have
  \begin{equation*}
    \lim_{n \to \infty} x_n = 0
  \end{equation*}
  if and only if,
  for every positive real number $\varepsilon$,
  there is a natural number $N$ such that,
  for every natural number $n > N$,
  we have $|x_n - 0| < \varepsilon$, i.e. $|x_n| < \varepsilon$.
  Let $\varepsilon \in \mathbb{R}$ such that $\varepsilon > 0$.
  It remains to prove that there is a suitable $N$.
  Necessarily, there is a $N \in \mathbb{N}$
  such that $N > 1 / \varepsilon$.
  It remains to prove that this $N$ is suitable.
  To that end, let $n \in \mathbb{N}$ such that $n > N$.
  Obviously, $n$ is positive.
  It remains to prove that
  $|  x_n| < \varepsilon$, i.e.
  $|1 / n| < \varepsilon$.
  Since $1 / \varepsilon < N$ and $N < n$,
  we have the inequality $1 / \varepsilon < n$.
  Certainly, each side of the inequality is positive.
  Thus,
  \begin{equation*}
    \begin{split}
      (1 / \varepsilon)^{-1} & > n^{-1} \\
                 \varepsilon & > 1 / n.
    \end{split}
  \end{equation*}
  Also, since $n$ is positive, we have $1 / n = | 1 / n |$.
  Thus, $| 1 / n | < \varepsilon$.
  QED

REPLY [2 votes]: Your proof is correct, but it should avoid the preamble, and can be shortened and made more rigorous by using:

Theorem (Archimedian property of the reals): Given real $x,y$ with $x>0$, there exists a natural number $N$ such that $Nx>y$.

Then the brief proof would be:

Given $\epsilon>0$, by the Archimedian property, there exists a natural number $N$ such that $N\epsilon>1$. Then if $n\geq N$, then $0<\frac{1}n\leq\frac{1}{N}<\epsilon$.
   Hence$|\frac{1}{n}-0|<\epsilon$ for $n\geq N$.
So by definition:
  $$\lim_{n\to\infty} \left|\frac{1}{n}\right| = 0$$