Does $f$ uniformly continuous on $[0, infty)$ imply that $lim_{xtoinfty} f(x)$ exists?












1












$begingroup$


Does $f$ uniformly continuous on $[0, infty)$ imply that $lim_{xtoinfty} f(x)$ exists?



I think that this is true. Because $f$ u.c. implies $f$ has bounded derivative. Can someone help me confirm this hypothesis?



Thanks






real-analysis limits







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    $sin x{{{}}}$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 14 '18 at 17:44










  • $begingroup$
    oh that is a counterexample. oops thanks
    $endgroup$
    – stackofhay42
    Dec 14 '18 at 17:45












  • $begingroup$
    wait so then if $f$ is continuous on $[0, infty)$ AND the $lim_{xtoinfty} f(x) = a$ for some $a in mathbb{R}$, then is it true that $f$ is uniformly continuous? this must be true then, right?
    $endgroup$
    – stackofhay42
    Dec 14 '18 at 17:46










  • $begingroup$
    Yes that is true stack.
    $endgroup$
    – zhw.
    Dec 14 '18 at 17:48
















1












$begingroup$


Does $f$ uniformly continuous on $[0, infty)$ imply that $lim_{xtoinfty} f(x)$ exists?



I think that this is true. Because $f$ u.c. implies $f$ has bounded derivative. Can someone help me confirm this hypothesis?



Thanks






real-analysis limits







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    $sin x{{{}}}$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 14 '18 at 17:44










  • $begingroup$
    oh that is a counterexample. oops thanks
    $endgroup$
    – stackofhay42
    Dec 14 '18 at 17:45












  • $begingroup$
    wait so then if $f$ is continuous on $[0, infty)$ AND the $lim_{xtoinfty} f(x) = a$ for some $a in mathbb{R}$, then is it true that $f$ is uniformly continuous? this must be true then, right?
    $endgroup$
    – stackofhay42
    Dec 14 '18 at 17:46










  • $begingroup$
    Yes that is true stack.
    $endgroup$
    – zhw.
    Dec 14 '18 at 17:48














1












1








1





$begingroup$


Does $f$ uniformly continuous on $[0, infty)$ imply that $lim_{xtoinfty} f(x)$ exists?



I think that this is true. Because $f$ u.c. implies $f$ has bounded derivative. Can someone help me confirm this hypothesis?



Thanks






real-analysis limits







share|cite|improve this question









$endgroup$




Does $f$ uniformly continuous on $[0, infty)$ imply that $lim_{xtoinfty} f(x)$ exists?



I think that this is true. Because $f$ u.c. implies $f$ has bounded derivative. Can someone help me confirm this hypothesis?



Thanks






real-analysis limits




real-analysis limits






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 14 '18 at 17:42









stackofhay42stackofhay42

2177




2177








  • 3




    $begingroup$
    $sin x{{{}}}$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 14 '18 at 17:44










  • $begingroup$
    oh that is a counterexample. oops thanks
    $endgroup$
    – stackofhay42
    Dec 14 '18 at 17:45












  • $begingroup$
    wait so then if $f$ is continuous on $[0, infty)$ AND the $lim_{xtoinfty} f(x) = a$ for some $a in mathbb{R}$, then is it true that $f$ is uniformly continuous? this must be true then, right?
    $endgroup$
    – stackofhay42
    Dec 14 '18 at 17:46










  • $begingroup$
    Yes that is true stack.
    $endgroup$
    – zhw.
    Dec 14 '18 at 17:48














  • 3




    $begingroup$
    $sin x{{{}}}$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 14 '18 at 17:44










  • $begingroup$
    oh that is a counterexample. oops thanks
    $endgroup$
    – stackofhay42
    Dec 14 '18 at 17:45












  • $begingroup$
    wait so then if $f$ is continuous on $[0, infty)$ AND the $lim_{xtoinfty} f(x) = a$ for some $a in mathbb{R}$, then is it true that $f$ is uniformly continuous? this must be true then, right?
    $endgroup$
    – stackofhay42
    Dec 14 '18 at 17:46










  • $begingroup$
    Yes that is true stack.
    $endgroup$
    – zhw.
    Dec 14 '18 at 17:48








3




3




$begingroup$
$sin x{{{}}}$?
$endgroup$
– Lord Shark the Unknown
Dec 14 '18 at 17:44




$begingroup$
$sin x{{{}}}$?
$endgroup$
– Lord Shark the Unknown
Dec 14 '18 at 17:44












$begingroup$
oh that is a counterexample. oops thanks
$endgroup$
– stackofhay42
Dec 14 '18 at 17:45






$begingroup$
oh that is a counterexample. oops thanks
$endgroup$
– stackofhay42
Dec 14 '18 at 17:45














$begingroup$
wait so then if $f$ is continuous on $[0, infty)$ AND the $lim_{xtoinfty} f(x) = a$ for some $a in mathbb{R}$, then is it true that $f$ is uniformly continuous? this must be true then, right?
$endgroup$
– stackofhay42
Dec 14 '18 at 17:46




$begingroup$
wait so then if $f$ is continuous on $[0, infty)$ AND the $lim_{xtoinfty} f(x) = a$ for some $a in mathbb{R}$, then is it true that $f$ is uniformly continuous? this must be true then, right?
$endgroup$
– stackofhay42
Dec 14 '18 at 17:46












$begingroup$
Yes that is true stack.
$endgroup$
– zhw.
Dec 14 '18 at 17:48




$begingroup$
Yes that is true stack.
$endgroup$
– zhw.
Dec 14 '18 at 17:48










2 Answers
2






active

oldest

votes


















1












$begingroup$

No—consider the function $sin(x)$.



It does not converge because it oscillates at infinity, but it clearly is uniformly continuous, being differentiable everywhere and having bounded derivative.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    so then if $f$ is continuous on $[0, infty)$ AND the $lim_{xtoinfty} f(x) = a$ for some $a in mathbb{R}$, then is it true that $f$ is uniformly continuous? this must be true then, right?
    $endgroup$
    – stackofhay42
    Dec 14 '18 at 17:47










  • $begingroup$
    @stackofhay42 Sorry, this is not related to your question directly but I've seen your trend in the past of never accepting an answer to any of your question. If an answer is complete and clear your doubt you should consider accepting it so that your question will be marked as answered.
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 17:51



















0












$begingroup$

The function $$f : xmapsto sqrt{x}$$



is uniformly continuous at $[0,+infty)$ but $$lim_{xto+infty}f(x)=+infty$$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    No—consider the function $sin(x)$.



    It does not converge because it oscillates at infinity, but it clearly is uniformly continuous, being differentiable everywhere and having bounded derivative.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      so then if $f$ is continuous on $[0, infty)$ AND the $lim_{xtoinfty} f(x) = a$ for some $a in mathbb{R}$, then is it true that $f$ is uniformly continuous? this must be true then, right?
      $endgroup$
      – stackofhay42
      Dec 14 '18 at 17:47










    • $begingroup$
      @stackofhay42 Sorry, this is not related to your question directly but I've seen your trend in the past of never accepting an answer to any of your question. If an answer is complete and clear your doubt you should consider accepting it so that your question will be marked as answered.
      $endgroup$
      – BigbearZzz
      Dec 14 '18 at 17:51
















    1












    $begingroup$

    No—consider the function $sin(x)$.



    It does not converge because it oscillates at infinity, but it clearly is uniformly continuous, being differentiable everywhere and having bounded derivative.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      so then if $f$ is continuous on $[0, infty)$ AND the $lim_{xtoinfty} f(x) = a$ for some $a in mathbb{R}$, then is it true that $f$ is uniformly continuous? this must be true then, right?
      $endgroup$
      – stackofhay42
      Dec 14 '18 at 17:47










    • $begingroup$
      @stackofhay42 Sorry, this is not related to your question directly but I've seen your trend in the past of never accepting an answer to any of your question. If an answer is complete and clear your doubt you should consider accepting it so that your question will be marked as answered.
      $endgroup$
      – BigbearZzz
      Dec 14 '18 at 17:51














    1












    1








    1





    $begingroup$

    No—consider the function $sin(x)$.



    It does not converge because it oscillates at infinity, but it clearly is uniformly continuous, being differentiable everywhere and having bounded derivative.






    share|cite|improve this answer









    $endgroup$



    No—consider the function $sin(x)$.



    It does not converge because it oscillates at infinity, but it clearly is uniformly continuous, being differentiable everywhere and having bounded derivative.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 14 '18 at 17:46









    AlexAlex

    1773




    1773












    • $begingroup$
      so then if $f$ is continuous on $[0, infty)$ AND the $lim_{xtoinfty} f(x) = a$ for some $a in mathbb{R}$, then is it true that $f$ is uniformly continuous? this must be true then, right?
      $endgroup$
      – stackofhay42
      Dec 14 '18 at 17:47










    • $begingroup$
      @stackofhay42 Sorry, this is not related to your question directly but I've seen your trend in the past of never accepting an answer to any of your question. If an answer is complete and clear your doubt you should consider accepting it so that your question will be marked as answered.
      $endgroup$
      – BigbearZzz
      Dec 14 '18 at 17:51


















    • $begingroup$
      so then if $f$ is continuous on $[0, infty)$ AND the $lim_{xtoinfty} f(x) = a$ for some $a in mathbb{R}$, then is it true that $f$ is uniformly continuous? this must be true then, right?
      $endgroup$
      – stackofhay42
      Dec 14 '18 at 17:47










    • $begingroup$
      @stackofhay42 Sorry, this is not related to your question directly but I've seen your trend in the past of never accepting an answer to any of your question. If an answer is complete and clear your doubt you should consider accepting it so that your question will be marked as answered.
      $endgroup$
      – BigbearZzz
      Dec 14 '18 at 17:51
















    $begingroup$
    so then if $f$ is continuous on $[0, infty)$ AND the $lim_{xtoinfty} f(x) = a$ for some $a in mathbb{R}$, then is it true that $f$ is uniformly continuous? this must be true then, right?
    $endgroup$
    – stackofhay42
    Dec 14 '18 at 17:47




    $begingroup$
    so then if $f$ is continuous on $[0, infty)$ AND the $lim_{xtoinfty} f(x) = a$ for some $a in mathbb{R}$, then is it true that $f$ is uniformly continuous? this must be true then, right?
    $endgroup$
    – stackofhay42
    Dec 14 '18 at 17:47












    $begingroup$
    @stackofhay42 Sorry, this is not related to your question directly but I've seen your trend in the past of never accepting an answer to any of your question. If an answer is complete and clear your doubt you should consider accepting it so that your question will be marked as answered.
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 17:51




    $begingroup$
    @stackofhay42 Sorry, this is not related to your question directly but I've seen your trend in the past of never accepting an answer to any of your question. If an answer is complete and clear your doubt you should consider accepting it so that your question will be marked as answered.
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 17:51











    0












    $begingroup$

    The function $$f : xmapsto sqrt{x}$$



    is uniformly continuous at $[0,+infty)$ but $$lim_{xto+infty}f(x)=+infty$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The function $$f : xmapsto sqrt{x}$$



      is uniformly continuous at $[0,+infty)$ but $$lim_{xto+infty}f(x)=+infty$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The function $$f : xmapsto sqrt{x}$$



        is uniformly continuous at $[0,+infty)$ but $$lim_{xto+infty}f(x)=+infty$$






        share|cite|improve this answer









        $endgroup$



        The function $$f : xmapsto sqrt{x}$$



        is uniformly continuous at $[0,+infty)$ but $$lim_{xto+infty}f(x)=+infty$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 14 '18 at 18:30









        hamam_Abdallahhamam_Abdallah

        38.1k21634




        38.1k21634






























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