- #1
wglmb
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Hi guys, I've been trying to do this for a while but I'm not really getting anywhere. Hints would be much appreciated!
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].
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!
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!