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Prove continuouty of 1/x (delta-epsilon)

  1. Oct 19, 2009 #1
    Hi guys, I've been trying to do this for a while but I'm not really getting anywhere. Hints would be much appreciated!

    1. The problem statement, all variables and given/known data

    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].


    2. Relevant equations



    3. 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!
     
  2. jcsd
  3. Oct 19, 2009 #2

    Landau

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    Science Advisor

    It is not a good idea to split cases. You will have to take a different delta to handle the situation at once. The thing that gives some trouble, is the denominator |ab|. We want to make sure this is small. Now, a is fixed (since we want to prove continuity at the fixed point a), so in fact the only trouble is 1/|b|, which we want to make small. This is the same as |b| being not too big. This can be arranged by choosing the right delta, which will depend on a. In that way, |a-b|<delta implies that b will be close to a, and after some fiddling you will be able to estimate 1/|b|.
    I hope this helps you. You can also take a look at this thread.
     
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