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Thread: evenly continues question

  1. #1
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    evenly continues question

    f is continues in (a,b)
    $\displaystyle lim_{x->b^-}f(x_0)=-\infty$
    prove that f is not "evenly continues"
    ??
    i dont know the proper term.

    basicly we need to prove that there is $\displaystyle \epsilon >0$ so for every $\displaystyle \delta >0$ there is x for which $\displaystyle |y_0-x_0|<\delta$ and $\displaystyle |f(y_0)-f(x_0)|>=\epsilon$

    if we choose $\displaystyle \epsilon=1$
    also we can use the limit
    ?
    Last edited by transgalactic; Aug 12th 2011 at 10:09 AM.
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  2. #2
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    Re: uniformly continuous question

    Quote Originally Posted by transgalactic View Post
    f is continues in (a,b)
    $\displaystyle lim_{x->b^-}f(x_0)=-\infty$
    prove that f is not "evenly continuous"
    ??
    i don't know the proper term.

    basicly we need to prove that there is $\displaystyle \epsilon >0$ so for every $\displaystyle \delta >0$ there is x for which $\displaystyle |x-x_0|<\delta$ and $\displaystyle |f-f(x_0)|>=\epsilon$

    if we choose $\displaystyle \epsilon=1$
    also we can use the limit
    ?
    Do you mean: 'uniformly' continuous ?
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  3. #3
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    Re: evenly continues question

    yes
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  4. #4
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    Re: evenly continues question

    Quote Originally Posted by transgalactic View Post
    f is continues in (a,b) $\displaystyle \lim_{x->b^-}f(x)=-\infty$
    prove that f is not "'uniformly' continuous"???
    Do you know this theorem: If f is uniformly' continuous on a set S and $\displaystyle (s_n)$ is a Cauchy Sequence in S then $\displaystyle f(s_n)$ is a Cauchy Sequence ?
    You can use proof by contradiction. There is a positive integer $\displaystyle K$ such that $\displaystyle b-\frac{1}{K}\in (a,b)$.
    Define $\displaystyle b_n=b-\frac{1}{K+n}$. Now apply the theorem.
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