3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ You can use epsilon-N to show it's always true $\endgroup$ – Vim Mar 2 '15 at 11:27
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

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

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

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

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

$\endgroup$
  • $\begingroup$ How do you always get a positive lower bound for $|x_n|$, what if $\{x_n\}\subset [-3,-1]$?? $\endgroup$ – Harry Mar 3 '15 at 7:24
  • $\begingroup$ @Harry If $\;\{x_n\}\subset [-3,-1]\;$ then $\;\forall\;n\,,\,\,-3<x_n<-1\implies |x_n|<3\;$ , right? $\endgroup$ – Timbuc Mar 3 '15 at 8:19
  • $\begingroup$ 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. $\endgroup$ – Harry Mar 3 '15 at 8:25
  • $\begingroup$ @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. $\endgroup$ – Timbuc Mar 3 '15 at 8:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.