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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?

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  • $\begingroup$ Since $\epsilon$ is arbitrary, you are correct that either way is technically correct. $\endgroup$ – Matt A Pelto Jan 15 '18 at 1:55
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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.

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  • $\begingroup$ Thank you. The rest of the statement is correct, also? Including the part about $|x_n| \geq M$ ? $\endgroup$ – jpugliese Jan 14 '18 at 18:07
  • $\begingroup$ I've seem your edit. For my, in principle, the part you highlighted was also troublesome. But I've seem the statement in more than one proof. By the course I'm taking right now, I can't use continuity to prove this... But we know that for all $\epsilon$, there exists $N \in \mathbb{N}$, all $n \geq N$, $|x_n - x|$ < $\epsilon$. Choosing $\epsilon$ equal to $|x|$ provides that $x_n$ must be different from 0 for $n \geq N_x$. Then, we get the existence of M > 0 s.t. $x_n$ must be greater or equal than $M$ from this $N_x$ onwards. Does that make sense for you? I'm also trying to figure this. $\endgroup$ – jpugliese Jan 14 '18 at 18:29
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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}

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  • $\begingroup$ The first half shows that $x_n \neq 0$, for all $n \geq N_o$. A statement like Timbuc's is true. We just either have to make the same assumption as Timbuc or check for instances of zero amongst the terms in the remaining finite sequence $\{x_j\}_{j=1}^{N_o-1}$. In either case we are discarding from the sequence $\{x_n\}_{n=1}^\infty$ at most finitely many instances of a "zero term", and we are left with at most finitely many distances from zero to choose the minimum from to have the statement "$|x_n|>M$ for all $n \in \mathbb{N}$". $\endgroup$ – Matt A Pelto Jan 15 '18 at 3:18
  • $\begingroup$ The aforementioned headache is avoided with "Select $N=\max \{N_o, N_1\}$." $\endgroup$ – Matt A Pelto Jan 15 '18 at 3:22
  • $\begingroup$ Thank you Matt. This made things clearer. Can you just please explain the triangle inequality here? $|0-x_n|+|x_n-x| \geq |0-x|$ I have difficulties with triangle inequalities. $\endgroup$ – jpugliese Jan 15 '18 at 13:43
  • $\begingroup$ proofwiki.org/wiki/Definition:Metric_Space/Distance_Function The standard euclidean distance $d$ between two real numbers $x$ and $y$ is given by $d(x,y):=|x-y|$. Notice $(M2)$ on proof wiki - aka the triangle inequality - is universally quantified for every $x,y,z \in \mathbb{R}$ which includes any term of the sequence $\{x_n\}_{n=1}^\infty$ or $0 \in \mathbb{R}$. $\endgroup$ – Matt A Pelto Jan 15 '18 at 18:53
  • $\begingroup$ I guess you might be more used to seeing this form associated with the term "triangle inequality" $|0-x_n|+|x_n-x| \geq |(0-x_n)+(x_n-x)|$ $\endgroup$ – Matt A Pelto Jan 15 '18 at 21:29

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