Proof: reciprocal of sequence converges to reciprocal of limit of sequence. Is it ok to have epsilon multiplying a constant? I want to prove that $$ lim_{n\to\infty}(1/x_n) = 1/x$$
I've read this, and I understand the proof by user Timbuc.
Convergence of Inverse of Convergent Sequence
I want to know if this statement is correct:
"Since $x_n$ converges to $x \neq 0$, there exists an $M \in R^+$ such that $|x_n|\geq M$ for all $n$ > $N$ (I get this from the fact that $x_n$ can't be zero but for a finite number of elements. Consider $N$ the number from which onwards this is valid). I can write:
$$|\frac{1}{x_n}-\frac{1}{x}| = |\frac{x}{x_nx} - \frac{x_n}{x_nx}|   = |\frac{x-x_n}{x_nx}| < \frac{\epsilon}{|x|M}   \space (1)$$ 
As $\frac{1}{|x|M}$ is a constant, this implies that $|\frac{1}{x_n}-\frac{1}{x}|$ converges since you can choose $\epsilon$ arbitrarily small."
More generally: the definition normally implies that $|x_n - x|< \epsilon$. User Timbuc, in his proof in the above link, manipulates so that he defines convergence as $|x_n - x|< \epsilon|x|M$ so that in the end he is left with just $\epsilon$.
If I end up with an arbitrary $\epsilon$ multiplying a constant, does (1) also implies convergence because I can pick $\epsilon$ arbitrarily small?
 A: Yes because the definition implies that you can choose $\epsilon$. Hence instead of choosing $\epsilon>0$ I choose $\displaystyle \frac{\epsilon}{\left|x\right|M}$ and then it will be majorated by $\epsilon$.
-EDIT FOR OP
For me it is not. I will make it ( for me ) clearer.
$\bullet$ You know that $\displaystyle x_n \underset{n \rightarrow +\infty}{\rightarrow}x$ so by definition for all $\epsilon_1>0$, it exists $N$ such as if $n \geq N$ hence $\left|x_n-x\right|<\epsilon_1$.
$\bullet$ Furthermore, $\displaystyle x \mapsto \frac{1}{x}$ is continuous in $x_n$ for $n \in \mathbb{N}$. Hence for all $\epsilon_2>0$, it exists $\eta>0$ such as if $\left|x-x_n\right|<\eta$ then $$\displaystyle\left|\frac{1}{x}-\frac{1}{x_n}\right|<\epsilon_2$$
The first point shows that for $n \geq N$ then $\eta=\epsilon_1$ and hence we have the result.
What perturbs me in your statement is 

"there exists $M>0$ such as $\left|x_n\right|\geq M$."

Because we know that the sequence $\left(x_n\right)_{n \in \mathbb{N}}$ converges then it is bounded so it would be $\left|x_n\right|<M$ and hence the following proof is false. But maybe i'm wrong.
A: To eventually bound the terms of the sequence away from zero:
Since $x \neq 0$, we know that $\frac{1}{2}|x|>0$. Since $\lim_{n\to \infty} x_n = x$, for this positive number $\frac{1}{2}|x|$ there is $N_o \in \mathbb{N}$ such that
\begin{equation} |x_n - x| < \frac{1}{2}|x| \,\text{ whenever }\, n\geq N_o.
\end{equation}
The triangle inequality gives us $|0-x_n|+|x_n-x| \geq |0-x|$, and so we have
\begin{aligned}|x_n| & \geq |x| - |x_n-x| 
\\& >|x| - \frac{1}{2}|x| \\&=\frac{1}{2}|x| \,\text{ for all }\, n\geq N_o.
\end{aligned}

Let $\varepsilon>0$ be given. Select $\varepsilon' = \min\{\varepsilon, \frac{1}{2}|x|^2\varepsilon\}$.  Since the sequence $\{x_n\}_{n=1}^\infty$ converges to $x$, for this positive number $\varepsilon'$ there is $N_1 \in \mathbb{N}$ such that
\begin{equation} |x_n - x| < \varepsilon' \,\text{ whenever }\, n\geq N_1.
\end{equation}
Select $N=\max\{N_o, N_1\}$. So if $n \geq N$, then
\begin{aligned}\left|\frac{1}{x_n}-\frac{1}{x}\right| & = \left|\frac{x}{x_nx} - \frac{x_n}{x_n x}\right|  \\& = \left|\frac{x-x_n}{x_nx}\right| \\& < \frac{2\varepsilon'}{|x|^2}
\\& \leq \varepsilon.
\end{aligned}
