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

?

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 ?

Re: evenly continues question

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.