Prove that $ frac{1}{x^5} $ is not uniformly continuous on $(0,2)$

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Prove that $frac{1}{x^5} $ is not uniformly continuous on $(0,2)$. I tried by putting $epsilon = 1$ but can't seem to find $x$ and $x'$ that would work.










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  • $(0,2)$...................
    – Nosrati
    Nov 15 at 19:48






  • 3




    A related exercise which might be interesting: prove that if $f : (a,b) to mathbb{R}$ is uniformly continuous, then $lim_{xto a^+} f(x)$ and $lim_{xto b^-} f(x)$ both exist.
    – Daniel Schepler
    Nov 15 at 20:01










  • This is a very good comment. Thanks a lot.
    – hamam_Abdallah
    Nov 16 at 18:02















up vote
1
down vote

favorite












Prove that $frac{1}{x^5} $ is not uniformly continuous on $(0,2)$. I tried by putting $epsilon = 1$ but can't seem to find $x$ and $x'$ that would work.










share|cite|improve this question
























  • $(0,2)$...................
    – Nosrati
    Nov 15 at 19:48






  • 3




    A related exercise which might be interesting: prove that if $f : (a,b) to mathbb{R}$ is uniformly continuous, then $lim_{xto a^+} f(x)$ and $lim_{xto b^-} f(x)$ both exist.
    – Daniel Schepler
    Nov 15 at 20:01










  • This is a very good comment. Thanks a lot.
    – hamam_Abdallah
    Nov 16 at 18:02













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Prove that $frac{1}{x^5} $ is not uniformly continuous on $(0,2)$. I tried by putting $epsilon = 1$ but can't seem to find $x$ and $x'$ that would work.










share|cite|improve this question















Prove that $frac{1}{x^5} $ is not uniformly continuous on $(0,2)$. I tried by putting $epsilon = 1$ but can't seem to find $x$ and $x'$ that would work.







calculus






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edited Nov 15 at 19:50









Ernie060

2,520319




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asked Nov 15 at 19:44









user15269

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  • $(0,2)$...................
    – Nosrati
    Nov 15 at 19:48






  • 3




    A related exercise which might be interesting: prove that if $f : (a,b) to mathbb{R}$ is uniformly continuous, then $lim_{xto a^+} f(x)$ and $lim_{xto b^-} f(x)$ both exist.
    – Daniel Schepler
    Nov 15 at 20:01










  • This is a very good comment. Thanks a lot.
    – hamam_Abdallah
    Nov 16 at 18:02


















  • $(0,2)$...................
    – Nosrati
    Nov 15 at 19:48






  • 3




    A related exercise which might be interesting: prove that if $f : (a,b) to mathbb{R}$ is uniformly continuous, then $lim_{xto a^+} f(x)$ and $lim_{xto b^-} f(x)$ both exist.
    – Daniel Schepler
    Nov 15 at 20:01










  • This is a very good comment. Thanks a lot.
    – hamam_Abdallah
    Nov 16 at 18:02
















$(0,2)$...................
– Nosrati
Nov 15 at 19:48




$(0,2)$...................
– Nosrati
Nov 15 at 19:48




3




3




A related exercise which might be interesting: prove that if $f : (a,b) to mathbb{R}$ is uniformly continuous, then $lim_{xto a^+} f(x)$ and $lim_{xto b^-} f(x)$ both exist.
– Daniel Schepler
Nov 15 at 20:01




A related exercise which might be interesting: prove that if $f : (a,b) to mathbb{R}$ is uniformly continuous, then $lim_{xto a^+} f(x)$ and $lim_{xto b^-} f(x)$ both exist.
– Daniel Schepler
Nov 15 at 20:01












This is a very good comment. Thanks a lot.
– hamam_Abdallah
Nov 16 at 18:02




This is a very good comment. Thanks a lot.
– hamam_Abdallah
Nov 16 at 18:02










1 Answer
1






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up vote
2
down vote



accepted










Let $$x_n=frac{1}{n}$$
and
$$y_n=frac{1}{n+1}$$



be two sequences of reals in $(0,2)$.



it is clear that
$$lim_{nto+infty}(y_n-x_n)=0$$



but



$$f(y_n)-f(x_n)=(n+1)^5-n^5$$
and
$$lim_{nto+infty}(f(y_n)-f(x_n))=+infty (ne 0)$$



We can conclude that $f$ is not uniformly continuous at $(0,2)$.



With $epsilon$.



Take $epsilon=1$ for example.
then for each $eta>0$,



and very very large $n$, we have



$$|y_n-x_n|<eta$$



because
$$lim_{nto+infty}(y_n-x_n)=0$$
and
$$|f(y_n)-f(x_n)|>1$$
because
$$lim_{nto+infty}(f(y_n)-f(x_n))=+infty$$






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    1 Answer
    1






    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote



    accepted










    Let $$x_n=frac{1}{n}$$
    and
    $$y_n=frac{1}{n+1}$$



    be two sequences of reals in $(0,2)$.



    it is clear that
    $$lim_{nto+infty}(y_n-x_n)=0$$



    but



    $$f(y_n)-f(x_n)=(n+1)^5-n^5$$
    and
    $$lim_{nto+infty}(f(y_n)-f(x_n))=+infty (ne 0)$$



    We can conclude that $f$ is not uniformly continuous at $(0,2)$.



    With $epsilon$.



    Take $epsilon=1$ for example.
    then for each $eta>0$,



    and very very large $n$, we have



    $$|y_n-x_n|<eta$$



    because
    $$lim_{nto+infty}(y_n-x_n)=0$$
    and
    $$|f(y_n)-f(x_n)|>1$$
    because
    $$lim_{nto+infty}(f(y_n)-f(x_n))=+infty$$






    share|cite|improve this answer



























      up vote
      2
      down vote



      accepted










      Let $$x_n=frac{1}{n}$$
      and
      $$y_n=frac{1}{n+1}$$



      be two sequences of reals in $(0,2)$.



      it is clear that
      $$lim_{nto+infty}(y_n-x_n)=0$$



      but



      $$f(y_n)-f(x_n)=(n+1)^5-n^5$$
      and
      $$lim_{nto+infty}(f(y_n)-f(x_n))=+infty (ne 0)$$



      We can conclude that $f$ is not uniformly continuous at $(0,2)$.



      With $epsilon$.



      Take $epsilon=1$ for example.
      then for each $eta>0$,



      and very very large $n$, we have



      $$|y_n-x_n|<eta$$



      because
      $$lim_{nto+infty}(y_n-x_n)=0$$
      and
      $$|f(y_n)-f(x_n)|>1$$
      because
      $$lim_{nto+infty}(f(y_n)-f(x_n))=+infty$$






      share|cite|improve this answer

























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        Let $$x_n=frac{1}{n}$$
        and
        $$y_n=frac{1}{n+1}$$



        be two sequences of reals in $(0,2)$.



        it is clear that
        $$lim_{nto+infty}(y_n-x_n)=0$$



        but



        $$f(y_n)-f(x_n)=(n+1)^5-n^5$$
        and
        $$lim_{nto+infty}(f(y_n)-f(x_n))=+infty (ne 0)$$



        We can conclude that $f$ is not uniformly continuous at $(0,2)$.



        With $epsilon$.



        Take $epsilon=1$ for example.
        then for each $eta>0$,



        and very very large $n$, we have



        $$|y_n-x_n|<eta$$



        because
        $$lim_{nto+infty}(y_n-x_n)=0$$
        and
        $$|f(y_n)-f(x_n)|>1$$
        because
        $$lim_{nto+infty}(f(y_n)-f(x_n))=+infty$$






        share|cite|improve this answer














        Let $$x_n=frac{1}{n}$$
        and
        $$y_n=frac{1}{n+1}$$



        be two sequences of reals in $(0,2)$.



        it is clear that
        $$lim_{nto+infty}(y_n-x_n)=0$$



        but



        $$f(y_n)-f(x_n)=(n+1)^5-n^5$$
        and
        $$lim_{nto+infty}(f(y_n)-f(x_n))=+infty (ne 0)$$



        We can conclude that $f$ is not uniformly continuous at $(0,2)$.



        With $epsilon$.



        Take $epsilon=1$ for example.
        then for each $eta>0$,



        and very very large $n$, we have



        $$|y_n-x_n|<eta$$



        because
        $$lim_{nto+infty}(y_n-x_n)=0$$
        and
        $$|f(y_n)-f(x_n)|>1$$
        because
        $$lim_{nto+infty}(f(y_n)-f(x_n))=+infty$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 16 at 18:01

























        answered Nov 15 at 19:50









        hamam_Abdallah

        37k21533




        37k21533






























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