# evenly continues question

• Aug 12th 2011, 09:06 AM
transgalactic
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
?
• Aug 12th 2011, 10:06 AM
SammyS
Re: uniformly continuous question
Quote:

Originally Posted by transgalactic
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 ?
• Aug 12th 2011, 10:08 AM
transgalactic
Re: evenly continues question
yes
• Aug 12th 2011, 02:39 PM
Plato
Re: evenly continues question
Quote:

Originally Posted by transgalactic
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.