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Math Help - Proving continuity using the epsilon-delta criterion - question 2

  1. #1
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    Proving continuity using the epsilon-delta criterion - question 2

    I figure the best way to see if i'm understanding this is to try it on another example:

    Show that the function 0, \infty) \rightarrow \mathbb{R}" alt="f0, \infty) \rightarrow \mathbb{R}" /> given by f(x)=\frac{1}{x} is continuous using the usual \epsilon-\delta definition.
    (this is the following question).

    \forall \epsilon>0 \exists \delta_{\epsilon}: \forall x \in E: |x-x_0|< \delta_{\epsilon} \Rightarrow |f(x)-f(x_0)|< \epsilon

    In this case, |f(x)-f(x_0)| < \epsilon is |\frac{1}{x}-\frac{1}{x_0}| < \epsilon.

    Starting from |x-x_0|<\delta_{\epsilon} \leq 1

    \frac{|x-x_0|}{|-xx_0|} < \frac{\delta_{\epsilon}}{|-xx_0|} where x,x_0 \neq 0 (which is acceptable since \frac{1}{x} \neq 0)

    \frac{|x_0-x|}{|xx_0|}< \frac{\delta_{\epsilon}}{|xx_0|}

    |\frac{1}{x}-\frac{1}{x_0}| < \frac{\delta_{\epsilon}}{|xx_0|}

    Let \delta=min \left\{ 1, \frac{\epsilon}{|xx_0|} \right \} for any given \epsilon>0.
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  2. #2
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    You cannot use x in the definition of \delta_{\epsilon} but only \epsilon and x_0
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    <br />
|x-x_0|<\delta_{\epsilon} \leq 1<br />

    x_0-1<x<1+x_0

    Therefore:

    <br />
|\frac{1}{x}-\frac{1}{x_0}| < \frac{\delta_{\epsilon}}{|xx_0|}<\frac{\delta_{\ep  silon}}{|(x_0-1)x_0|}<br />

    Let <br />
\delta=min \left\{ 1, \frac{\epsilon}{|(x_0-1)x_0|} \right \}<br />
for any given \epsilon>0.
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    First, because this is a new problem you should have started a new thread.

    What you have done above has nothing to do with the correct proof.
    Conceptually this particular type of problem is one of the most difficult.
    As you did note if s \in \left( {0,1} \right)\; \Rightarrow \;s \ne 0.
    Now we must do some preliminary work. \frac{s}{2} > 0 so we can say that
    \left| {x - s} \right| < \frac{s}{2}\; \Rightarrow \; - \frac{s}{2} < x - s < \frac{s}<br />
{2}\; \Rightarrow \;\frac{s}{2} < x.
    Because we are working with denominators we have \frac{1}{x} < \frac{2}{s}.

    Follow this very closely. If \varepsilon  > 0 then pick <br />
\delta  = \min \left\{ {\frac{s}{2},\frac{{\varepsilon s^2 }}{2}} \right\} so that
    \left| {x - s} \right| < \delta \, \Rightarrow \,\left| {\frac{1}{x} - \frac{1}<br />
{s}} \right| = \frac{{\left| {s - x} \right|}}{{xs}} < \frac{2}{s}\frac{\delta }<br />
{s} = \frac{{2s^2 \varepsilon }}{{s^2 2}} = \varepsilon .
    Last edited by Plato; January 18th 2009 at 02:02 PM.
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