# Convergence of Inverse of Convergent Sequence

Let $\{x_n\}$ be a sequence in $\mathbb{R}$ where $\forall n\in\mathbb{N}:x_n\neq 0$ and it converges to some $x\neq 0$. If the sequence is NOT monotone is it ever true that $\frac{1}{x_n}\rightarrow\frac{1}{x}$? If so what other conditions (if any) are needed and how would you show it.

Thanks in advance any feedback is greatly appreciated.

• You can use epsilon-N to show it's always true – Vim Mar 2 '15 at 11:27

It is always true, provided that $x\neq 0$ and each $x_n\neq 0$.

Generally, $f(x_n)\to f(x)$ if $f$ is continuous and $x$ and each $x_n$ are in the domain of $f$. This follows from continuity.

Consider the mapping: $g: \mathbb{R}^+\to \mathbb{R}^+, \quad x \mapsto \frac{1}{x}$. $g$ is continuous in its domain. This implies that if $x_n \to x$ then $g(x_n) \to g(x)$, that is your thesis.

Same argument can be used in the case of negative values.

EDIT: As MPW wrote it is sufficient to take $g: \mathbb{R} \setminus \{0\}\to \mathbb{R} \setminus \{0\}$ without splitting cases.

• Sign doesn't matter. Just take domain and range to be $\mathbb R\setminus \{0\}$. – MPW Mar 2 '15 at 11:31
• @MPW You are right. I splitted the two cases for no reason – EmarJ Mar 2 '15 at 11:33

Suppose $$\;x_n\to x\neq 0\;$$ , and $$\;x_n\neq 0\;$$ (enough to assume for almost all $$\;n\in\Bbb N\;$$ and then throw away all the zero terms).

Now there exists $$\;M\in\Bbb R^+\;$$ s.t. $$\;|x_n|\ge M\;\;\forall\,n\in\Bbb N\;$$ (why?), and then for all $$\;\epsilon >0\;$$ there exists $$\;K\in\Bbb N\;$$ s.t. that for all

$$\;n>K\;,\;\;|x_n-x|<|x|\epsilon M\;\;\implies$$

$$\implies\;\left|\frac1{x_n}-\frac1x\right|=\left|\frac{x-x_n}{xx_n}\right|<\frac{|x|\epsilon M}{|x|M}=\epsilon$$

• How do you always get a positive lower bound for $|x_n|$, what if $\{x_n\}\subset [-3,-1]$?? – Harry Mar 3 '15 at 7:24
• @Harry If $\;\{x_n\}\subset [-3,-1]\;$ then $\;\forall\;n\,,\,\,-3<x_n<-1\implies |x_n|<3\;$ , right? – Timbuc Mar 3 '15 at 8:19
• Yes, 3 is an upper bound of $|x_n|$, and the problem is still there as in your proof above you need a lower bound of $|x_n|$ which is positive, you say a $M$, s.t.$|x_n|\geq M$ for all $n$. So that's what I don't understand, can you clear this up for me? Thanks. – Harry Mar 3 '15 at 8:25
• @Harry: $$|x_n|\ge M\implies\frac1{|x_n|}\le\frac1M$$ which is what was used in the last inequality. I added some editing for clearity to my answer. – Timbuc Mar 3 '15 at 8:47