If $f$ is continuous and with a limit at infinity then $f$ is uniformly continuous












0












$begingroup$



Let $f : [a, +infty) rightarrow mathbb{R}$ be continuous, such that
$lim_{x rightarrow +infty} f(x) = l in mathbb{R}$. Show that f is
uniformly continuous.




I want to split the domain - one in which we have continuity, and the other in which the values of f are approaching $l$. This means there are theree possibilities for our $x$ and $y$ using the uniformly continuous definition: x and y are in the first part, x and y are in the second, x is in the first and y in the second.



How would I do this proof?






real-analysis limits continuity uniform-continuity







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    To avoid splitting into three cases, you could allow the pieces to overlap and make sure to choose $delta$ less than the width of the overlap. This avoids the case where $x$ is in one piece and $y$ is in the other.
    $endgroup$
    – J. Loreaux
    Nov 7 '16 at 19:06












  • $begingroup$
    @GregordeCillia A uniformly continuous function can have unbounded derivative
    $endgroup$
    – zhw.
    Nov 7 '16 at 19:12












  • $begingroup$
    Yes. That's why I deleted my comment after a few mjnutes. I was thinking anout lipschitz continuouity and wanted to give an osciallator-like counterexample
    $endgroup$
    – Gregor de Cillia
    Nov 7 '16 at 19:21










  • $begingroup$
    Fix epsilon. Let delta(x) be the maximum delta used in the definition of continuity at x. Show delta(x) has a nonzero limit given the conditions of the problem.
    $endgroup$
    – Alephnull
    Nov 7 '16 at 20:35
















0












$begingroup$



Let $f : [a, +infty) rightarrow mathbb{R}$ be continuous, such that
$lim_{x rightarrow +infty} f(x) = l in mathbb{R}$. Show that f is
uniformly continuous.




I want to split the domain - one in which we have continuity, and the other in which the values of f are approaching $l$. This means there are theree possibilities for our $x$ and $y$ using the uniformly continuous definition: x and y are in the first part, x and y are in the second, x is in the first and y in the second.



How would I do this proof?






real-analysis limits continuity uniform-continuity







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    To avoid splitting into three cases, you could allow the pieces to overlap and make sure to choose $delta$ less than the width of the overlap. This avoids the case where $x$ is in one piece and $y$ is in the other.
    $endgroup$
    – J. Loreaux
    Nov 7 '16 at 19:06












  • $begingroup$
    @GregordeCillia A uniformly continuous function can have unbounded derivative
    $endgroup$
    – zhw.
    Nov 7 '16 at 19:12












  • $begingroup$
    Yes. That's why I deleted my comment after a few mjnutes. I was thinking anout lipschitz continuouity and wanted to give an osciallator-like counterexample
    $endgroup$
    – Gregor de Cillia
    Nov 7 '16 at 19:21










  • $begingroup$
    Fix epsilon. Let delta(x) be the maximum delta used in the definition of continuity at x. Show delta(x) has a nonzero limit given the conditions of the problem.
    $endgroup$
    – Alephnull
    Nov 7 '16 at 20:35














0












0








0





$begingroup$



Let $f : [a, +infty) rightarrow mathbb{R}$ be continuous, such that
$lim_{x rightarrow +infty} f(x) = l in mathbb{R}$. Show that f is
uniformly continuous.




I want to split the domain - one in which we have continuity, and the other in which the values of f are approaching $l$. This means there are theree possibilities for our $x$ and $y$ using the uniformly continuous definition: x and y are in the first part, x and y are in the second, x is in the first and y in the second.



How would I do this proof?






real-analysis limits continuity uniform-continuity







share|cite|improve this question











$endgroup$





Let $f : [a, +infty) rightarrow mathbb{R}$ be continuous, such that
$lim_{x rightarrow +infty} f(x) = l in mathbb{R}$. Show that f is
uniformly continuous.




I want to split the domain - one in which we have continuity, and the other in which the values of f are approaching $l$. This means there are theree possibilities for our $x$ and $y$ using the uniformly continuous definition: x and y are in the first part, x and y are in the second, x is in the first and y in the second.



How would I do this proof?






real-analysis limits continuity uniform-continuity




real-analysis limits continuity uniform-continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 7 '16 at 20:16









Did

248k23224462




248k23224462










asked Nov 7 '16 at 19:01









user381164user381164

724




724








  • 2




    $begingroup$
    To avoid splitting into three cases, you could allow the pieces to overlap and make sure to choose $delta$ less than the width of the overlap. This avoids the case where $x$ is in one piece and $y$ is in the other.
    $endgroup$
    – J. Loreaux
    Nov 7 '16 at 19:06












  • $begingroup$
    @GregordeCillia A uniformly continuous function can have unbounded derivative
    $endgroup$
    – zhw.
    Nov 7 '16 at 19:12












  • $begingroup$
    Yes. That's why I deleted my comment after a few mjnutes. I was thinking anout lipschitz continuouity and wanted to give an osciallator-like counterexample
    $endgroup$
    – Gregor de Cillia
    Nov 7 '16 at 19:21










  • $begingroup$
    Fix epsilon. Let delta(x) be the maximum delta used in the definition of continuity at x. Show delta(x) has a nonzero limit given the conditions of the problem.
    $endgroup$
    – Alephnull
    Nov 7 '16 at 20:35














  • 2




    $begingroup$
    To avoid splitting into three cases, you could allow the pieces to overlap and make sure to choose $delta$ less than the width of the overlap. This avoids the case where $x$ is in one piece and $y$ is in the other.
    $endgroup$
    – J. Loreaux
    Nov 7 '16 at 19:06












  • $begingroup$
    @GregordeCillia A uniformly continuous function can have unbounded derivative
    $endgroup$
    – zhw.
    Nov 7 '16 at 19:12












  • $begingroup$
    Yes. That's why I deleted my comment after a few mjnutes. I was thinking anout lipschitz continuouity and wanted to give an osciallator-like counterexample
    $endgroup$
    – Gregor de Cillia
    Nov 7 '16 at 19:21










  • $begingroup$
    Fix epsilon. Let delta(x) be the maximum delta used in the definition of continuity at x. Show delta(x) has a nonzero limit given the conditions of the problem.
    $endgroup$
    – Alephnull
    Nov 7 '16 at 20:35








2




2




$begingroup$
To avoid splitting into three cases, you could allow the pieces to overlap and make sure to choose $delta$ less than the width of the overlap. This avoids the case where $x$ is in one piece and $y$ is in the other.
$endgroup$
– J. Loreaux
Nov 7 '16 at 19:06






$begingroup$
To avoid splitting into three cases, you could allow the pieces to overlap and make sure to choose $delta$ less than the width of the overlap. This avoids the case where $x$ is in one piece and $y$ is in the other.
$endgroup$
– J. Loreaux
Nov 7 '16 at 19:06














$begingroup$
@GregordeCillia A uniformly continuous function can have unbounded derivative
$endgroup$
– zhw.
Nov 7 '16 at 19:12






$begingroup$
@GregordeCillia A uniformly continuous function can have unbounded derivative
$endgroup$
– zhw.
Nov 7 '16 at 19:12














$begingroup$
Yes. That's why I deleted my comment after a few mjnutes. I was thinking anout lipschitz continuouity and wanted to give an osciallator-like counterexample
$endgroup$
– Gregor de Cillia
Nov 7 '16 at 19:21




$begingroup$
Yes. That's why I deleted my comment after a few mjnutes. I was thinking anout lipschitz continuouity and wanted to give an osciallator-like counterexample
$endgroup$
– Gregor de Cillia
Nov 7 '16 at 19:21












$begingroup$
Fix epsilon. Let delta(x) be the maximum delta used in the definition of continuity at x. Show delta(x) has a nonzero limit given the conditions of the problem.
$endgroup$
– Alephnull
Nov 7 '16 at 20:35




$begingroup$
Fix epsilon. Let delta(x) be the maximum delta used in the definition of continuity at x. Show delta(x) has a nonzero limit given the conditions of the problem.
$endgroup$
– Alephnull
Nov 7 '16 at 20:35










1 Answer
1






active

oldest

votes


















4












$begingroup$

It suffices you do the following. Given $epsilon>0$, take $N$ large so that, for $x>N$,



$$|f(x)- l|<varepsilon/2$$



It follows that if $x,y>N$, then $|f(x)-f(y)|<varepsilon$. Now consider the interval $[a,N+1]$, which is compact. Here $f$ is uniformly continuous, so there is some $delta >0$ such that if $|x-y|<delta$ then $|f(x)-f(y)|<varepsilon$. We can assume that $delta<1$.



Consider now arbitrary $x,ygeqslant a$ that are at most $delta$ units apart. If both $x,yleqslant N+1$, then you're done. Our choice of $delta$ is such that if $x> N+1$ then also $y>N$, and you're also done.






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2003989%2fif-f-is-continuous-and-with-a-limit-at-infinity-then-f-is-uniformly-continuo%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    It suffices you do the following. Given $epsilon>0$, take $N$ large so that, for $x>N$,



    $$|f(x)- l|<varepsilon/2$$



    It follows that if $x,y>N$, then $|f(x)-f(y)|<varepsilon$. Now consider the interval $[a,N+1]$, which is compact. Here $f$ is uniformly continuous, so there is some $delta >0$ such that if $|x-y|<delta$ then $|f(x)-f(y)|<varepsilon$. We can assume that $delta<1$.



    Consider now arbitrary $x,ygeqslant a$ that are at most $delta$ units apart. If both $x,yleqslant N+1$, then you're done. Our choice of $delta$ is such that if $x> N+1$ then also $y>N$, and you're also done.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      It suffices you do the following. Given $epsilon>0$, take $N$ large so that, for $x>N$,



      $$|f(x)- l|<varepsilon/2$$



      It follows that if $x,y>N$, then $|f(x)-f(y)|<varepsilon$. Now consider the interval $[a,N+1]$, which is compact. Here $f$ is uniformly continuous, so there is some $delta >0$ such that if $|x-y|<delta$ then $|f(x)-f(y)|<varepsilon$. We can assume that $delta<1$.



      Consider now arbitrary $x,ygeqslant a$ that are at most $delta$ units apart. If both $x,yleqslant N+1$, then you're done. Our choice of $delta$ is such that if $x> N+1$ then also $y>N$, and you're also done.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        It suffices you do the following. Given $epsilon>0$, take $N$ large so that, for $x>N$,



        $$|f(x)- l|<varepsilon/2$$



        It follows that if $x,y>N$, then $|f(x)-f(y)|<varepsilon$. Now consider the interval $[a,N+1]$, which is compact. Here $f$ is uniformly continuous, so there is some $delta >0$ such that if $|x-y|<delta$ then $|f(x)-f(y)|<varepsilon$. We can assume that $delta<1$.



        Consider now arbitrary $x,ygeqslant a$ that are at most $delta$ units apart. If both $x,yleqslant N+1$, then you're done. Our choice of $delta$ is such that if $x> N+1$ then also $y>N$, and you're also done.






        share|cite|improve this answer









        $endgroup$



        It suffices you do the following. Given $epsilon>0$, take $N$ large so that, for $x>N$,



        $$|f(x)- l|<varepsilon/2$$



        It follows that if $x,y>N$, then $|f(x)-f(y)|<varepsilon$. Now consider the interval $[a,N+1]$, which is compact. Here $f$ is uniformly continuous, so there is some $delta >0$ such that if $|x-y|<delta$ then $|f(x)-f(y)|<varepsilon$. We can assume that $delta<1$.



        Consider now arbitrary $x,ygeqslant a$ that are at most $delta$ units apart. If both $x,yleqslant N+1$, then you're done. Our choice of $delta$ is such that if $x> N+1$ then also $y>N$, and you're also done.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 7 '16 at 19:13









        Pedro Tamaroff♦Pedro Tamaroff

        96.9k10153297




        96.9k10153297






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2003989%2fif-f-is-continuous-and-with-a-limit-at-infinity-then-f-is-uniformly-continuo%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
            Read more

            Puebla de Zaragoza

            Image
            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
            Read more

            Musa

            Image
            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
            Read more