- #1

- 17

- 0

## Homework Statement

Prove that the function [itex]g(x)=1/x[/itex] is continuous on [itex]\latexbb{R}\smallsetminus\{0\}[/itex], but cannot be defined at the origin [itex]0[/itex] in such a way that the resulting function is continuous on [itex]\latexbb{R}[/itex].

## Homework Equations

## The Attempt at a Solution

Just the first part for now - proving continuity on [itex]\latexbb{R}\smallsetminus\{0\}[/itex].

They want an [itex]\epsilon-\delta[/itex] proof. Here we go...

Let [itex]a,b\in\latexbb{R}\smallsetminus\{0\}[/itex].

Let [itex]\delta=\epsilon>0[/itex].

Then [itex]\forall a\in\latexbb{R}\smallsetminus\{0\}[/itex] and [itex]\forall\epsilon>0[/itex] we have:

[itex]\mid a-b\mid<\delta\Rightarrow\mid g(a)-g(b)\mid=\mid \frac{1}{a}-\frac{1}{b}\mid=\mid\frac{b-a}{ab}\mid<\frac{\delta}{\mid ab\mid}[/itex]

Now I take the case where [itex]b[/itex] is such that [itex]\mid a-b\mid>1[/itex] and we have...

[itex]\frac{\delta}{\mid ab\mid}<\delta=\epsilon[/itex]

So it's proven for such [itex]b[/itex] (I think?)

Now I have no idea what to do about [itex]b[/itex] such that [itex]\mid a-b\mid<1[/itex]. Or was splitting it into two cases a bad idea? Am I going anywhere useful here?

Please help!