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Math Help - Most likely an easy continuity question

  1. #1
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    Most likely an easy continuity question

    Hi everyone, this is probably an easy question but I'm having trouble on the wording of the proof.

    Let f be continuous at x=c and f(c) > 1

    Show that there exists an r > 0 such that \forall x \in B(c,r) \bigcap D : f(x) > 1

    Like I said, probably easy, but I don't know how to show it on paper.
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  2. #2
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    Quote Originally Posted by HeirToPendragon View Post
    Hi everyone, this is probably an easy question but I'm having trouble on the wording of the proof.

    Let f be continuous at x=c and f(c) > 1

    Show that there exists an r > 0 such that \forall x \in B(c,r) \bigcap D : f(x) > 1

    Like I said, probably easy, but I don't know how to show it on paper.
    Use the definition of "continuity" taking \epsilon= \frac{f(c)}{2}.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    Use the definition of "continuity" taking \epsilon= \frac{f(c)}{2}.
    It may be that \frac{f(c)}{2} < 1

    To fix it consider: \varepsilon  = \frac{{f(c) - 1}}<br />
{2}\, \Rightarrow \,\left| {f(x) - f(c)} \right| < \frac{{f(c) - 1}}<br />
{2}\,\, \Rightarrow \,1 < \frac{{f(c) + 1}}<br />
{2} < f(x)\,.
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  4. #4
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    Quote Originally Posted by Plato View Post
    It may be that \frac{f(c)}{2} < 1

    To fix it consider: \varepsilon  = \frac{{f(c) - 1}}<br />
{2}\, \Rightarrow \,\left| {f(x) - f(c)} \right| < \frac{{f(c) - 1}}<br />
{2}\,\, \Rightarrow \,1 < \frac{{f(c) + 1}}<br />
{2} < f(x)\,.
    Yep, that works out perfectly. Thanks.
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  5. #5
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    Thanks, Plato. I was thinking of the more common problem, f(x)> 0, and didn't pay attention to the "> 1" part.
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